Download Interesting problems from the AMATYC Student Math League Exams

Interesting problems from the AMATYC Student Math League Exams
(November 2003, #1) If L has equation ax  by  c , M is its reflection across the y-axis, and N
is its reflection across the x-axis, which of the following must be true about M and N for all
nonzero choices of a, b, and c?
M
L
N
Suppose that L has an x-intercept of A and a y-intercept of B. Then its slope is 
an x-intercept of -A and a y-intercept of B. So its slope is 
of A and a y-intercept of -B,. So its slope is 
B
. M has
A
B B
 . N has an x-intercept
A A
B B
 .
A A
So the correct answer is C) the slopes are equal.
Or just notice that the addition of the dashed line must result in a parallelogram which forces
M and N to have the same slope.
Or notice that the equation of M must be a   x   by  c so its slope is
of N must be ax  b   y   c so its slope must be
a
, and the equation
b
a
.[See the section on Graph Properties]
b
(November 2003, #2) A collection is made up of an equal number of pennies, nickels, dimes,
and quarters. What is the largest possible value of the collection which is less than $2?
41T  200  T 
200
 T  4 . So the largest possible value is 4  41  $1.64 .
41
So the correct answer is A) $1.64.
(November 2003, #3) When the polynomial P  x  is divided by  x  2  , the remainder is
2
3x  3 . What is the remainder when  x  1 P  x  is divided by  x  1 x  2  ?
2
P  x   Q  x  x  2    3x  3 , so
 x  1 P  x   Q  x  x  1 x  2   x  1 3x  3 .
2
this means that the remainder when  x  1 P  x  is divided by  x  1 x  2  is
2
2
But
 x  1 3x  3  3x 2  6 x  3 .
So the correct answer is B) 3x 2  6 x  3 . [See the section on Polynomial Properties]


(November 2003, #4) If f  x   3x  2 , find f f  f  3  .


f f  f  3   f  f  7    f 19   55 .
So the correct answer is B) 55.
(November 2003, #5) What is the remainder when x3  2 x 2  4 is divided by x  2 ?
 2
3
 2  2   4  8  8  4  12 . So the answer is 12 .
2
So the correct answer is A) 12 . [See the section on Polynomial Properties]
(November 2003, #6) Let p be a prime number and k an integer such that x 2  kx  p  0 has
two positive integer solutions. What is the value of k  p ?
x 2  kx  p must factor into  x  a  x  b  where a and b are integers. But the only way this
can happen is if one of them is p and the other is 1.
This means that
x 2  kx  p   x  1 x  p   x 2   p  1 x  p , so k    p  1 .
This rearranges into
k  p  1.
So the correct answer is B) 1. [See the section on Polynomial Properties]
(November 2003, #7) What is the least number of prime numbers (not necessarily different)
that 3185 must be multiplied by so that the product is a perfect cube?
3185  5  637  5  7  91  5  7  7  13 . So to turn it into a perfect cube with the least number of
multiplications by primes, we would need to multiply it by 52  7  132 .
So the correct answer is E) 5. [See The Fundamental Theorem of Arithmetic]
(November 2003, #8) Two adjacent faces of a three-dimensional rectangular box have areas 24
and 36. If the length, width and height of the box are all integers, how many different volumes
are possible for the box?
H
L
LH  36 and WH  24 , and V  LWH 
W
 LH WH   36  24
, where H is a common factor of
H
H
36 and 24. So the number of different volumes is the same as the number of common factors of
36 and 24. The common factors are 1, 2, 3, 4, 6, and 12.
So the correct answer is E) 6.
(November 2003, #9)
tan t  sin t cos t

tan t
tan t  sin t cos t
 1  cos 2 t  sin 2 t .
tan t
So the correct answer is C) sin 2 t . [See the section on Trigonometric Formulas]
(November 2003, #10) The counting numbers are written in the pattern at the right. Find the
middle number of the 40th row.
1
2 3 4
5 6 7 8 9
10 11 12 13 14 15 16
The middle numbers are generated by adding consecutive even numbers to 1.
M 40  1  2  4  6 
 1  2 1  2  3 
1 2 
 2  39
 39
.
39  40
 1  39  40  1561
2
Or another pattern is that the middle number of the nth row is given by M n  n 2   n  1 , so
M 40  402  39  1561 .
So the correct answer is A) 1561. [See the section on Algebraic Formulas]
(November 2003, #11) The solution set of x 2  3x  18  0 is a subset of the solution set of
which of the following inequalities?
x 2  3x  18   x  6  x  3  0
S
S
S
-3
x  x  20   x  5  x  4   0
6
S
2
S
No
S
S
-4
5
x4
0
x3
S
No
S
S
S
-3
4
S
x2  8x  14  0
Yes
S
S
S
S
4 2
4 2
So the correct answer is C) x2  8x  14 .
(November 2003, #12) If 2a  4b  128b3  16a3 and a  2b , find a 2  2ab  b2 .
2a  4b  128b3  16a3  a  2b  8 8b3  a3   a  2b  8  2b  a   4b 2  2ab  a 2 
  a  2ab  4b
2
2

.
1
8
1
So the correct answer is A)  . [See the section on Algebraic Formulas]
8
(November 2003, #13) Square ABCD is inscribed in circle O, and its area is a. Square EFGH
H
is inscribed in a semicircle of O.
What is the area of square EFGH?
G
D
C
E
F
A
The area of an inscribed square in a circle of radius r, is 2r 2 .
B
r
2
r
45
2r
The area of an inscribed square in a semicircle of radius r, is
r
4r 2
.
5
s
s
2
So if a  2r 2 , then the area of the square inscribed inside the semicircle is
So the correct answer is B)
2a
.
5
2a
. [See the section on Geometric Formulas]
5
(November 2003, #14) Consider all arrangements of the letters AMATYC with either the A’s
together or the A’s on the ends. What fraction of all possible such arrangements satisfy these
conditions?
A’s together: 5!
A’s on the ends: 4!
Total number of different arrangements:
6!
2
5! 4! 2  4!1  5  2  4! 6 2


 .
6!
6!
6!
5
2
So the correct answer is D)
2
. [See the section on Sets and Counting]
5
(November 2003, #15) The year 2003 is prime, but its reversal, 3002, in not. In fact, 3002 is
the product of exactly three different primes. Let N be the sum of these three primes. How
many other positive integers are the products of exactly three different primes with this sum N?
3002  2  19  79 , so N  2  19  79  100 . p1  p2  p3  100 , since the sum of three distinct
primes not equal to 2 must be an odd number, one of the primes must be 2. p2  p3  98
Now let’s check the possible prime values for p2 , and see which ones make 98  p2 a prime
number as well.
p2  3,5,7,11,13,17,19,23,29, 31, 37,41, 43,47,53,59, 61, 67, 71,73, 79,83, 89, 97
So there are two other numbers: 2  31  67  4154 and 2  37  61  4514
So the correct answer is C) 2.
(November 2003, #16) In a group of 30 students, 25 are taking math, 22 English, and 19
history. If the largest and smallest number who could be taking all three courses are M and m
respectively, find M  m .
M  19
m   25  22  30  19  30  6
So M  m  19  6  25 .
So the correct answer is E) 25. [See the section on Sets and Counting]
(November 2003, #17) A boat with an ill passenger is 7½ miles north of a straight coastline
which runs east and west. A hospital on the coast is 60 miles from the point on shore south of
the boat. If the boat starts toward shore at 15 mph at the same time an ambulance leaves the
hospital at 60 mph and meets the ambulance, what is the total distance(to the nearest .5 mile)
traveled by the boat and the ambulance?
boat
15t
15
2
60  60t
60t
hospital
15t 
14625
7200  1575
2
 15 
     60  60t   3375t 2  7200t 
0t 
 .83 .
4
6750
2
2
2
So the total distance traveled is 75  .83  62.5 .
So the correct answer is E) 62.5.
AMATYC  MYM represents a different
(November 2003, #18) If each letter in the equation
decimal digit, find T’s value.
AMATYC   MYM  and to produce a 6-digit number, MYM must be at least 323. So let’s start
2
checking the squares of MYM values:
MYM  323,343,353, 363
3632  131769 , so T  7 .
So the correct answer is E) 7.
(November 2003, #19) If a, b, c, and d are nonzero numbers such that c and d are solutions of
x2  ax  b  0 and a and b are solutions of x 2  cx  d  0 , find a  b  c  d .
If c and d are solutions of x2  ax  b  0 , then
x 2  ax  b   x  c  x  d   x 2   c  d  x  cd , so c  d  a and cd  b .
If a and b are solutions of x 2  cx  d  0 , then
x 2  cx  d   x  a  x  b   x 2   a  b  x  ab , so a  b  c and ab  d .
So we get the system
acd 0
abc0
cd  b
ab  d
Solving the linear part
acd 0
abc0
Leads to
.
acd 0
bd 0
With solutions of d  t , b  t , c  s, a    s  t  ; s, t  0 .
Plugging this into the other two
equations leads to st  t  s  1 and t  s  t   t  t  2 . So the solution of the system is
a  1, b  2, c  1, d  2 , so a  b  c  d  1  2  1  2  2 .
So the correct answer is A) 2 .
(November 2003, #20) Al and Bob are at opposite ends of a diameter of a silo in the shape of a
tall right circular cylinder with radius 150 feet. Al is due west of Bob. Al begins walking along
the edge of the silo at 6 feet per second at the same moment that Bob begins to walk due east at
the same speed. The value closest to the time in seconds when Al first can see Bob is
 x,
150
2
 x2

 150 2 
,0 

 x




At a point on the circle where a tangent line will intersect the x-axis, x,
 150 2 
,0  .
intersect the x-axis at the point 
 x



Here’s why:
150
2

 x 2 , it will

The slope of the tangent line at the point on the circle x0 ,
an equation of the tangent line is y 
150 
2

 x  


2
0
150
x0
150 
2
2

 x02 is 
x0
150 
2
x
, so
2
0

  x  x  . Setting y equal
0
2 
 x0 
 150 2 
,0  .
to zero and solving for x to find the x-intercept results in 
 x0




t 

x

150cos





25 


;0  t  25 . So we need to solve the
Al’s path can be parametrized by 

t
 y  150sin    
25 


equation 150  6t 
Bob's position
150 
2
t 

150cos    
25 

.
An approximate solution of this equation is
where tangent line intersects
48.00747736.
Here’s the graph of both sides of the equation:
So the correct answer is C) 48.
(February 2004, #1) A stock loses 60% of its value. What must the percent of increase be to
recover all of its lost value?
.40 x 1  p   x  1  p 
1
1
 p   1  1.5  150% .
.4
.4
So the correct answer is C) 150%.
(February 2004, #2) Which of the following is NOT a factor of x 4  4 x3  x 2  16 x  12 ?
 1
4
 4  1   1  16  1  12  1  4  1  16  12  24  0 , so x  1 is not a factor of
3
2
x 4  4 x3  x 2  16 x  12 .
So the correct answer is D) x  1. [See the section on Polynomial Properties]
(February 2004, #3) The library in Johnson City has between 1000 and 2000 books. Of these,
1
1
25% are fiction,
are biographies, and
are atlases. How many books are either
13
17
biographies or atlases?
The total number of books must be a common multiple of 4, 13, and 17. The LCM of these
numbers is 884. The total number of books must be a multiple of the LCM that’s between 100
1768
0 and 2000, so it must be 1768. The number of biographies is
 136 , and the number of
13
1768
atlases is
 104 , so the number of books that are either biographies or atlases is 240.
17
So the correct answer is A) 240. [See the section on LCM and GCF]
(February 2004, #4) A tricimal is like a decimal, except the digits represent fractions with
16 1 2 1
77
powers of 3 instead of 10. For example,
  
 .121 as a tricimal. How is
27 3 9 27
81
expressed as a tricimal?
77 2 75
 
81 81 81
2 25
 
81 27
2
1 24
 

81 27 27
2
1 8
 

81 27 9
2
1 2 6
 
 
81 27 9 9
2
1 2 2
 
   .2212
81 27 9 3
So the correct answer is E) .2212.
(February 2004, #5) The function P  t   cos8t can be written as sums and differences of
powers of cost . When P  t  is written that way, what is the coefficient of  cost  ?
3
Since cos 2 x  2cos 2 x  1, we get that
2
2
2


cos8t  2cos 2 4t  1  2  2cos 2 2t  1  1  2 2  2cos 2 t  1  1  1 .





Expanding the
right side will only yield even powers of cost , so the coefficient of  cost  is zero.
3
So the correct answer is A) 0. [See the section on Trigonometric Formulas]
(February 2004, #6) If log a b  64 , find log a2 b3 .
log a b  64  b  a64  b3  a643  b3   a 2 
643
2
 b3   a 2   log a2 b3  96 .
96
So the correct answer is D) 96. [See the section on Logarithmic Properties]
(February 2004, #7) The number 877530p765q6 is divisible by both 8 and 11, with p and q
both digits from 0 to 9. The number is also divisible by
In order for the number to be divisible by 8, 5q6, the number consisting of the ones, tens, and
hundreds digits must be divisible by 8. This means that the only possible values for q are 3 and
7. In order for the number to be divisible by 11, the difference in the sum of the even digit
positions and the odd digit positions must be divisible by 11. For 877530p765q6 the difference
of these sums is 8  7  3  p  6  q    7  5  0  7  5  6   p  q  6 . So the only
possibility is q  3, p  3 , which leads to the number 877530376536. Since the sum of its digits
is 8  7  7  5  3  0  3  7  6  5  3  6  60 which is divisible by 3, the number must be
also be divisible by 3, 6, 12, 24, and 33.
So the correct answer is B) 12. [See the section on Divisibility Rules]
(February 2004, #8) Teams A and B play a series of games; whoever wins two games first
wins the series. If Team A has a 70% chance of winning any single game, what is the
probability that Team A wins the series?
Team A will be the winner only if the following results occur:
AA, BAA, ABA.
These occur with the following probabilities:
.7  , .3.7  , .3.7 
2
2
2
.
So the probability that Team A wins the series is .7   .3.7   .3.7   .7  1.6   .784 .
2
2
2
2
So the correct answer is E) .784. [See the section on Probability Formulas]
(February 2004, #9) The Venn diagram at the right represents sets A, B, and C(not necessarily
in that order). Depending on how the diagram is labeled, how many different answers are
possible for the number of elements in the set A  B ? (Note: A  B is all elements which are in
A but not in B.)
The possible designations of the sets A, B, and C are
A
B
A
C
B
C
B
C
A
C
A
B
2
0
2
B
C
C
B
A
A
These lead to the following values for the number of elements in A  B :
4, 8, 4, 5, 5, 2, so there are 4 different values.
So the correct answer is C) 4.
1
4
4
3
(February 2004, #10) A fixed point for a function y  f  x  is a real number r such that
f  r   r . How many of the following functions must have a fixed point?
A polynomial function of even degree > 0 without a fixed point would be f  x   x 2  x  1 .
A rational function f  x  
xa
x 1
without a fixed point would be f  x  
.
xb
x 1
For a polynomial function of odd degree > 1, f  x  . The equation f  x   x leads to the
polynomial equation f  x   x  0 . Since the left side is an odd degree polynomial, it must
have at least one real solution, and hence the original function must have a fixed point.
For a trigonometric function f  x   Asin Bx  D , the graph must intersect the graph of the
identity function, and hence it must have at least one fixed point.
So the correct answer is C) 2.
(February 2004, #11) Which of the following is the identity function f  x   x for all real
numbers?
eln x  x only for x  0 .
ln e x for all x.
sin  arcsin x   x only for 1  x  1 .
eln x  x only for x  0
arc tan  tan x   x only for 
So the correct answer is B) ln e x .

2
x

2
(February 2004, #12) A circular table is pushed into the corner of a rectangular room so that it
touches both walls. A point on the edge of the table between the two points of contact is 2
inches from one wall and 9 inches from the other wall. What is the radius of the table?
x2  y 2  r 2
r 2
r 9
So  r  9    r  2   r 2  r 2  22r  85  0   r  5 r  17   0 .
2
2
So the correct answer is D) 17 inches. [See the section on Geometric Formulas]
4
(February 2004, #13) In ABC , C  90 and cos A  . If D is the midpoint of side AC,
5
find cos CDB .
A
2
5
D
13
2
C
3
B
So cos CDB 
2
2 13
.

13
13
So the correct answer is A)
2 13
.
13
(February 2004, #14) Enrique walks along a level road and then up a hill. At the top of the hill
he immediately turns and walks back to his starting point. He walks 4 mph on level ground, 3
mph uphill, and 6 mph downhill. If the entire walk takes 6 hours, how far does he walk?
His walk consists of two equal level ground portions of length L, one uphill portion of length U,
and one equal downhill portion of length D. So the total distance he walks is equal to
2 L  U  D  2L  2U  2  L  U  .
We also know that
2L U U
   6  6 L  6U  72  2  L  U   24 .
4
3 6
So the correct answer is C) 24 mi..
(February 2004, #15) If x 2  x  3 , then x3 
x 2  x  3  x 3  x  x  3  x 3  x 2  3 x  x 3  x  3  3x  x 3  4 x  3 .
So the correct answer is B) 4 x  3 .
(February 2004, #16) A bag holds 5 cards identical except for color. Two are red on both
sides, two are black on both sides, and one is red on one side and black on the other. If you
pick a card at random and see that the only side you can see is red, what is the probability that
the other side is red?
An equally likely sample space for the experiment is
S   R, R  ,  R, R  ,  R, R  ,  R, R  ,  B, B  ,  B, B  ,  B, B ,  B, B ,  R, B ,  B, R  , where the first
part of the ordered pair is the color you see, and the second part is the color you don’t see.
If the color you see is red, then the sample space has been reduced to
 R, R  ,  R, R  ,  R, R  ,  R, R  ,  R, B  and of these, the only outcomes with the other side also
red are  R, R  ,  R, R  ,  R, R  ,  R, R  , so the probability is
So the correct answer is D)
4
.
5
4
. [See the section on Probability Formulas]
5
(February 2004, #17)The set S contains the number 2, and if it contains the number n, it also
contains 3n and n  5 (assume S contains only numbers produced by these rules). Which of the
following is NOT in S?
For a number to be in S, then some sequence of subtractions of 5 and divisions by 3 should take
you back to the number 2.
2001  3  667  5  665  5  662  5  657  3  219  3  73  5  68  5  63  3  21  3
752
2002  5  1997  5  1992  3  664  5  659  5  654  3  218  5  213  3  71  5
 66  3  22  5  17  5  12  5  7  5  2
2003  5  1998  3  666  3  222  3  74  5  69  3  23  5  18  3  6  3  2
2004  3  668  5  663  3  221  5  216  3  72  5  67  5  62  5  57  5
 52  5  47  5  42  5  37  5  32  5  27  5  22  5  17  5  12  5  7  5  2
This leaves 2000.
So the correct answer is A) 2000.
(February 2004, #18) Let f  x   ax  b , with b  a both positive integers. If for positive
integers p and q, f  p   18 and f  q   39 , what is the value of b?
f  p   18  ap  b  18 and f  q   39  aq  b  39 , so subtracting leads to a  q  p   21 .
From this we can conclude that a  1,3,7,21. Since b  a and both are positive integers, we can
disregard 1. If a  3 , then the possible values of b are 1 and 2. b  1 leads to 3 p  1  18 ,
which doesn’t yield an integer solution. b  2 leads to 3 p  2  18 , which doesn’t yield an
integer solution. If a  7 , then the possible values of b are 1, 2, 3, 4, 5, and 6. b  4 leads to
7 p  4  18 and 7q  4  39 which yield integer solutions. So b must be 4.
So the correct answer is C) 4.
(February 2004, #20) Ed has four children, Al, Bo, Cy, and Di(in order oldest to youngest).
Bo is 4 years older than Cy and 12 years older than Di. This year Ed notices that he is twice as
old as Bo, and the sum of the squares of the children’s ages equals the square of Ed’s age. If Di
just became a math teacher, what is the sum of the children’s ages?
B  C  4, B  D  12, E  2B, A2  B2  C 2  D2  E 2 , B  30, D  C  B  A . So we get that
A2  B 2   B  4    B  12    2 B   A2  B 2  32 B  160 .
2
2
2
Let’s check values of B
starting with 30 to see if the right side is a square and the other conditions are satisfied. The
first value to work is B  38 . It produces the values A  50, C  34, D  26, E  76 , and all the
other conditions are satisfied.
The sum of the children’s ages is
A  B  C  D  50  38  34  26  148 .
So the correct answer is E) 148.
(October 2004, #2) In square ABCD, point E is between A and B, and point F is between B and
C. Find the sum of the measures of AEF and EFC .
E
A
B
F
D
C
Consider the pentagon AEFCD. It can be dissected into three triangles with angle sum of 540 ,
and so the angle sum of the pentagon is also 540 . To get the sum of the angles AEF and
EFC , we just have to subtract the three 90 angles at A, D and C. So the sum of the angles
AEF and EFC is 540  270  270 .
So the correct answer is C) 270 . [See the section on Geometric Formulas]
(October 2004, #4) A newspaper advertises that it sells the Sunday paper for one-third the price
of the rest of the week’s papers. If a weekly subscription costs between $2.20 and $2.30, what
is the cost of one Sunday paper and one daily paper?
S  13  6 D  2 D . 220  6D  2D  230  27.5  D  28.75 . So a daily paper costs 28 cents
and a Sunday paper costs 56 cents, giving a total of 84 cents.
So the correct answer is C) $.84.
(October 2004, #6) A date is called weird if the number of its month and the number of its day
have greatest common factor 1. What are the fewest number of weird days in any month?
We want to look at months with numbers that have a lot of different factors: 6(June) and
12(December). It might also be nice to look at months with a fewer number of days.
6(June): 1, 5, 7, 11, 13, 17, 19, 23, 25, 29 for a total of 10 weird days.
12(December): 1, 5, 7, 11, 13, 17, 19, 23, 25, 29, 31 for a total of 11 weird days.
So the correct answer is B) 10.
(October 2004, #7) Lucia is not yet 80 years old. Each of her sons has as many sons as
brothers. The combined number of Lucia’s sons and grandsons equals her age, and her oldest
grandson is 29. How old is Lucia? Place your numerical answer in the corresponding answer
blank.
Let B be the number of Lucia’s sons, S be the number of Lucia’s grandsons, and L be Lucia’s
age. Then we get the following equations: S  B  1, B  BS  L  B2  L . So we know that
Lucia’s age must be a square number less than 80 and greater than 59. Lucia must be 64.
So the correct answer is 64.
5
5
5
4
(October 2004, #8) What is arc csc  arc sec  arc cot  arc cot ?
4
4
4
5
5
5
5
4
Let   arc csc ,  arc sec ,   arc cot ,   arc cot .
4
4
4
5

5
4
5



3
4
From the pictures, you can see that
5
5
5
4
 
arc csc  arc sec  arc cot  arc cot             .
4
4
4
5
2 2
So the correct answer is B)  .
(October 2004, #9) George bought groceries with a $10 bill. The cost of the groceries had
three different digits, and the amount of his change had the same 3 digits in a different order.
What was the sum of the digits in the cost?
a.bc
Suppose the cost was $a.bc and the change was $c.ab, then it must be that  c.ab .
So
10.00
c  b  10, b  a  9, a  c  9 , and hence 2  a  b  c   10  9  9  28 . So the sum of the digits
in the cost is 14.
So the correct answer is B) 14.
(October 2004, #10) Let N be the smallest number divisible by 33 which is greater than
1,000,000 and whose digits are all 0’s and 1’s. What are N’s leading four digits?
Assuming that N is a seven-digit number whose digits are all 0’s and 1’s that is divisible by 33,
then it must start with a 1 and be divisible by 3 and 11. To be divisible by 3, the sum of its
digits must be divisible by 3, which means that it has three 1’s or six 1’s. To be divisible by 11
means that the difference in alternate digit sums must be divisible by 11, which means that the
difference in alternate digit sums must be 0, so the number of 1’s must be even. Now we know
that it starts with a 1 and has 5 additional 1’s and one 0. So the 6 possibilities are
1011111, 1101111, 1110111, 1111011, 1111101, 1111110.
But the only ones divisible by 11 are 1101111, 1111011, and 1111110. The smallest one is
1101111.
So the correct answer is D) 1101. [See the section on Divisibility Rules]
(October 2004, #12) The song “What a Beautiful Life” has the lyric, “Day 18,253, well, honey,
that’s fifty years.” If the lyric was supposed to be exactly correct, by how many days is it
wrong?
Every fourth year is a leap year of 366 days, unless the year is divisible by 100 and doesn’t
leave a remainder of 200 or 600 when divided by 900. This gives the average year
approximately 365.24 days. 50  365.24  18,262 , so the lyric is off by about 9 days.
So the correct answer is D) 8 to 10.
(October 2004, #13) Chris traveled 1 hour longer and 2 miles farther than Calvin, but averaged
3 mph slower. If the sum of their times was 4 hours, what was the sum in miles of the distances
they traveled?
Tchris  Tcalvin  1 , Dchris  Dcalvin  2 ,
3
5
and Tchris  . Combining this with the second and third
2
2
 2 Dcalvin
57
. So the sum of the distances is
 3  3  Dcalvin 
4
2
last equations we get that Tcalvin 
equations leads to
Dcalvin
5
2
Dchris Dcalvin

 3 , and Tchris  Tcalvin  4 . From the first and
Tchris
Tcalvin
57 57
61

 2   30.5 .
4
4
2
So the correct answer is D) 30.5.
1
which is
3
the hypotenuse of a right triangular region in Quadrant I with legs the x and y axes and area
392
.
3
y
(October 2004, #15) Find the sum of the x and y intercepts of the line with slope 
x
1
392
392
implies that y  13 x . Area of
implies that 12 xy 
. Combining the
3
3
3
392
x 2 392
1 1
two equations leads to 2 x 3 x 
 
 x 2  784  x  28 . So the x-intercept is 28
3
6
3
28
112
and the y-intercept is
, giving a sum of
.
3
3
The slope of 
So the correct answer is D)
112
. [See the section on Equations of Lines]
3
[See the section on Geometric Formulas]
(October 2004, #16) Let A  0,1,2,3,4,5,6,7,8,9 . How many three-element subsets of A
contain at least two consecutive integers?
Let’s count how many sets of 3 don’t have any consecutive integers:
Smallest element
Other two elements
0
2,4
0
2,5
0
2,6
0
2,7
0
2,8
0
2,9
0
3,5
0
3,6
0
3,7
0
3,8
0
3,9
0
4,6
0
4,7
0
4,8
0
4,9
0
5,7
0
5,8
0
5,9
0
6,8
0
6,9
0
7,9
The number of three element sets with smallest element 0 without any consecutive integers is
1  2  3  4  5  6  21. Similarly, the number of three element sets with smallest element 1 is
1  2  3  4  5  15 . Similarly, the number of three element sets with smallest element 2 is
1  2  3  4  10 . Similarly, the number of three element sets with smallest element 3 is
1  2  3  6 . Similarly, the number of three element sets with smallest element 4 is 1  2  3 .
Similarly, the number of three element sets with smallest element 5 is 1 . So the number of
subsets of size 3 without any consecutive integers is 21  15  10  6  3  1  56 . Since the
10!
10  9  8

 120 , there must be 120  56  64
total number of subsets of size 3 is
10  3! 3! 3  2  1
subsets of size 3 that contain at least two consecutive integers.
So the correct answer is E) 64. [See the section on Sets and Counting]
(October 2004, #17) If x, y and z are positive integers with x  2 y  2 z  2005 and
2 x  2 y  z  2004 , find the smallest possible value of x  y  z .
Consider the system
x  2 y  2 z  2005
. Subtracting twice the first equation from the second
2 x  2 y  z  2004
x  2 y  2 z  2005
equation leads to
. Adding the second equation to the first equation leads to
2 y  3z  2006
x  z  1
2 y  3z  2006
.
Dividing the second equation by 2 leads to
x  z  1
y  32 z  1003
.
So the
solutions are given by z  t , y  1003  32 t , x  t  1 , where t is an even integer with 1  t 
This means that x  y  z  12 t  1002 , where t is an even integer with 1  t 
2006
.
3
2006
. So the
3
smallest possible value of x  y  z is 1003.
So the correct answer is E) 1003.
(October 2004, #18) A store has four open checkout stands. In how many ways could six
customers line up at the checkout stands?
Here are the possibilities sorted by largest number at a single checkout stand:
# of people
6 0 0 0 ways to line up # of arrangements Total
Checkout stand 1 2 3 4
# of people
4!
4
3!
2880
5 1 0 0 ways to line up # of arrangements Total
Checkout stand 1 2 3 4
# of people
6!  720
6  5!  720
4!
 12
2!
8640
4 1 1 0 ways to line up # of arrangements Total
Checkout stand 1 2 3 4 15  2  4!  720
4!
 12
2!
8640
# of people
4 2 0 0 ways to line up # of arrangements Total
Checkout stand 1 2 3 4 15  4! 2!  720
# of people
3 3 0 0
Checkout stand 1 2 3 4
# of people
3 2 1 0
3 1 1 1
20  3! 3!  720
ways to line up
ways to line up
Checkout stand 1 2 3 4 20  3  2  3!  720
# of people
2 2 2 0
ways to line up
Checkout stand 1 2 3 4 15  6  2! 2! 2!  720
# of people
2 2 1 1
8640
ways to line up # of arrangements Total
Checkout stand 1 2 3 4 20  3  3! 2!  720
# of people
4!
 12
2!
ways to line up
Checkout stand 1 2 3 4 15  6  2  2! 2!  720
4!
6
2! 2!
4320
# of arrangements Total
4!  24
17280
# of arrangements Total
4!
4
3!
2880
# of arrangements Total
4!
4
3!
2880
# of arrangements Total
4!
6
2! 2!
So the total number of ways the customers can line up is
2880  8640  8640  8640  4320  17280  2880  2880  4320  60480 .
So the correct answer is D) 60480. [See the section on Sets and Counting]
4320
(October 2004, #20) Suppose f  x   ax  b , g  x   bx  a (a, b integers). If f 1  8 and
f  g  50   g  f  50   28 , then find the product of a and b.
f 1  8  a  b  8 .
f  g  50    g  f  50    28   50ab  a 2  b    50ab  b 2  a   28  a 2  b 2   a  b   28
  a  b   a  b  1  28  7  a  b   28  a  b  4
So
a b8
ab4
.
, and this means that a  6 and b  2 . Therefore, ab  12 .
So the correct answer is B) 12. [See the section on Algebraic Formulas]
(February 2005, #5) A palindrome is a word or a number (like RADAR or 1221) which reads
the same forwards and backwards. If dates are written in the format MMDDYY, how many
dates in the 21st century are palindromes?
Let’s deal with the months that are one digit:
1
M(0)
9
M(1-9)
2
1
1
1
18
D(1,2)
D(same as
first day
digit)
Y(same as
second month
digit)
Y(same as
first month
digit)
Total
Now for the months that are two digit:
1
M(1)
3
M(0,1,2)
2
1
1
1
6
D(1,2)
D(same as
first day
digit)
Y(same as
second month
digit)
Y(same as
first month
digit)
Total
So the number of dates in the 21st century that are palindromes is 24.
So the correct answer is C) 24. [See the section on Sets and Counting]
(February 2005, #8) Mrs. Abbott finds that the number of possible groups of 3 students in her
class is exactly five times the number of possible groups of 2 students. How many students are
in her class?
The number of possible groups of 3 students is
n  n  1 n  2 
n!
.

6
 n  3! 3!
The
number
n  n  1 n  2 
6
of

possible
5n  n  1
2

groups
of
2
students
is
n  n  1
n!
.

2
 n  2 ! 2!
So
n2 5
  n  17 .
6
2
So the correct answer is B) 17. [See the section on Sets and Counting]
(February 2005, #9) In how many ways can slashes be placed among the letters
AMATYCSML to separate them into four groups with each group including at least one letter?
Three slashes must be placed in the spaces between the letters. Since there are 8 spaces, there
8!
876

 56 ways to do this.
are
6
8  3! 3!
So the correct answer is B) 56. [See the section on Sets and Counting]
(February 2005, #10) Two motorists set out at the same time to go from Danbury to Norwich,
100 miles apart. They follow the same route and travel at different but constant speeds of an
integral number of miles per hour. The difference in their speeds is a prime number of miles
per hour, and after driving for two hours, the distance of the slower car from Danbury is five
times that of the faster car from Norwich. What is the faster car’s speed?
rf  rs  p,2rs  5 100  2  rs  p   rs 
5  50  p 
. So we need to find a prime number, p,
6
so that 50  p is divisible by 6. p  2 does the job, and we get rs  40 and rf  42 .
So the correct answer is B) 42 mph.
(February 2005, #11) The sum cos1  cos 2  cos 3 
 cos 357  cos 358  cos 359 is
equal to
Rearranging the sum leads to
 cos1  cos181   cos 2  cos182   cos 3  cos183  
000
  cos179  cos 359   cos180
 0  1  1
So the correct answer is E) 1. [See the section on Trigonometric Formulas]
0 2 
0 5
N

(February 2005, #12) If M  
and
, find M 2005 .



5 0 
2 0
0 2  0 2  10 0 
1 0 
2005
2 1002
M

M
M  101002 M .
M2 



10
.









5 0  5 0   0 10
0 1 
So the correct answer is A) 101002 M . [See the section on Matrix Multiplication]
(February 2005, #13) A basketball team scores 78 points on 41 baskets(field goals count 2
points, free throws 1 point, and 3-point shots 3 points). If the number of each type of basket is
different, and the number of baskets of any of the three types differs by no more than 4, how
many field goals are scored?
Let B be the number of field goals, F the number of free throws, and T the number of 3-point
B  F  T  41
shots. Then we have the system
. Subtracting twice the first equation from
2 B  F  3T  78
B  F  T  41
the second equation leads to
. Adding the second equation to the first equation
 F  T  4
B  2T  37
leads to
. So we get that B  37  2T , F  T  4,T  T . The number of free throws
 F  T  4
is always 4 more than the number of 3-point shots.
We also need to have
41
T  4  37  2T  T  4  11  T  . So the possible values of T are 12 and 13, but only 12
3
works. This leads to B  37  2  12  13 .
So the correct answer is C) 13.
(February 2005, #14) Which of the following is a factor of
10
2005
 1  102005  2   102005  ?
2
2
2
Let x  102005 .
Then
10
2005
 1  102005  2   102005    x  1   x  2   x 2  x 2  6 x  5   x  1 x  5  .
2
2
2
2
So the factors are 102005  1 and 102005  5  .
So the correct answer is D) 102005  5  .
2
(February 2005, #15) The volume of cylinder A is 108 , which is twice the volume of
cylinder B. If the radius and height of A are the height and radius respectively of B, find the
height of cylinder B.
 rA2 hA  108 , rB2 hB  54 , rA  hB , hA  rB  rB2 hB  54, hB2 rB  108  hB  2rB  2rB3  54
 r  27  rB  3
3
B
.
So the correct answer is A) 3.
(February 2005, #17) Two triangular regions are formed in the first quadrant, one with vertices
 0,0  ,  5,0  , and  0,12  , the other with vertices  0,0  , 8,0  , and  0,6  . Find the area to the
nearest integer of the region they have in common.
12
x y
 1
5 12
6
36
11
x y
 1
8 6
5
40
11
8
The common area can be calculated as the sum of the areas of a trapezoid and a triangle.
1
2
 19 111 .
 6  1136   1140  12  1511  1136  1151  1140  1511  1811  2310
11
2
So the nearest integer is 19.
So the correct answer is C) 19. [See the section on Geometric Formulas]
(February 2005, #18) A triangle has sides of length a, b, and c which are consecutive integers
5
in increasing order, and cos C  . Find cos A .
16
From the Law of Cosines, we get
5
 n 2  4n  4  2n 2  2n  1  85 n 2  85 n
16
.
2
3 2
21
 8 n  8 n  3  0  n  7n  8  0   n  8 n  1  0  n  8
 n  2
2
 n2   n  1  2n  n  1 
2
Again, from the Law of Cosines, we get
8  1  8  2   82

cos A 
2  8  18  2 
2
2
So the correct answer is C)

181  64 117 13

 .
180
180 20
13
. [See the section on Trigonometric Formulas]
20
(February 2005, #19) If p  5 is a prime number, what is the largest integer which must be a
factor of p 4  1 ?
p4  1
2400
120
150
180
240
400
Y
Y
N
Y
Y
14640
Y
N
Y
N
So the answer is either 120 and 240. If p 4  1 is always divisible by 240, then it will always be
divisible by 120 as well. If we can show that p 4  1 always has at least four factors of 2, then
the answer must be 240. p 4  1   p  1 p  1  p 2  1 .
Both
 p  1
and
 p  1
are
consecutive even integers and hence there product must be divisible by 8, and  p 2  1 must be
even, so  p  1 p  1  p 2  1 must have 16 as a factor, so the answer is 240.
So the correct answer is D) 240.
(October 2005, #6) Let M and L be two perpendicular lines tangent to a circle with radius 6.
Find the area bounded by the two lines and the circle.
The area is the difference between the area of the square of side 6 and the quarter circle of
radius 6. So it’s 36  14  36  36  9 .
6
So the correct answer is B) 36  9 .
6
[See the section on Geometric Formulas]
(October 2005, #7) When I am as old as my father is now, I will be five times as old as my son
is now. By then, my son will be eight years older than I am now. The sum of my father’s age
and my age is 100 years. How much older am I than my son?
F  5S , S   F  M   8  M , F  M  100 . These lead to 16S  208  S  13, F  65, M  35 .
So M  S  35  13  22 .
So the correct answer is D) 22 yrs..
(October 2005, #9) If a 2  b2  33 and a3  b3  817 have integer solutions with a  b , find
the value of a  b .
a 2  b 2  33   a  b  a  b   33  a  b  1,3,11.
a 3  b3  817   a  b   a 2  ab  b 2   817 which means that a  b would have to be a factor
of 817. Since 3 and 11 are not factors of 817, a  b would have to be 1.
So the correct answer is A) 1. [See the section on Algebraic Formulas]
(October 2005, #10) SML has sides of length 6, 7, and 8.
 cos S  cos M  cos L  .
Find the exact value of
51
96
21
From the Law of Cosines: 64  36  49  84cos M  cos M 
84
77
36  64  49  112cos L  cos L 
112
49  36  64  96cos S  cos S 
So  cos S  cos M  cos L  
So the correct answer is B)
51 21 77 47


 .
96 84 112 32
47
. [See the section on Trigonometric Formulas]
32
(October 2005, #11) Find the sum of all the solutions of cos x  cot x cos x for which
0  x  2 .
cos2 x
 sin x cos x  cos2 x  0  cos x  sin x  cos x   0 .
sin x
 3  5
solutions are 2 , 2 , 4 , 4 , and the sum of the solutions is 2  32  4  54  72  3.5 .
cos x  cot x cos x  cos x 
So the
So the correct answer is C) 3.5 . [See the section on Trigonometric Formulas]


(October 2005, #13) For i  1,2,3,4,5,6 , let log a logb  log c  xi    0 , where a, b, and c
represent every possible different arrangement of 2, 4, and 8. The product x1 x2 x3 x4 x5 x6 can be
expressed in the form 2 N . Find N.


log a logb  logc  xi    0  logb  logc  xi    1  logc  xi   b  xi  cb . The arrangements of
2, 4, and 8 are the following: 248,284,428,482,824,842
These lead to the following values:
x1  84  212 , x2  48  216 , x3  82  26 , x4  28 , x5  42  24 , x6  24 , so x1 x2 x3 x4 x5 x6  250 .
So the correct answer is E) 50. [See the section on Logarithmic Properties]
(October 2005, #14) A triangle has vertices A  0,0  , B  3,0  , and C  3,4  . If the triangle is
rotated counterclockwise around the origin until C lies on the positive y-axis, find the area of
the intersection of the region bounded by
the original triangle and the region
bounded by the rotated triangle.
The intersection is a right triangle whose
area is 32 h .
  tan 1 43   2  tan 1 43   2 tan 1 43  2 ,
so h  3tan  2 tan 1 43  2  .
h

3
This means that the area of the intersection is
9 tan  2 tan 1 43  2 
2
9 

 
So the correct answer is A)

 
 
cot tan 1 43 1
2 cot tan 1 43
2

2

9cot  2 tan 1 34 
2
1
9  162 3  21
  4
2
16
9
21
. [See the section on Geometric Formulas]
16
[See the section on Trigonometric Formulas]
(October 2005, #15) When written as a decimal number, 20052005 has D digits and leading digit
L. Find D  L .
In general, the number of decimal digits of a number N is log N   1, where  x  is the greatest
integer less than or equal to x. Since log  20052005   2005log  2005   6620.74 , the number of
digits
in

the
log 20052005

number


is
6620  1  6621 .
We
also
have
that
log 20052005 6620
20052005  10
 10
 106620  100.73932  106620  5.4868  106620 , so the leading
digit is 5. So D  L  6621  5  6626 .
So the correct answer is D) 6626.
(October 2005, #16) if 0  t  2 , 0  z  1, and cos t 
true? z 
1  z2
, how many of the following are
1  z2
1  cos t
2z
2z
t
,sin t 
,
tan
t

,
z

tan
1  cos t
1  z2
1  z2
2
1  z2
2z
2z
1  z2
1  cos t
cos t 
 sin t 
 tan t 


and cos t 
2
2
2
2
1 z
1 z
1 z
1 z
1  cos t
2sin 2 2t
1  cos t
t
, so all four are true.
z


tan
1  cos t
2cos 2 2t
2
So the correct answer is E) 4. [See the section on Trigonometric Formulas]
2 z2
1 z 2
2
1 z 2
 z2  z
(October 2005, #17) Let a1  2 and an1 
12
for all n  1 . Find the value that an
2an  5
approaches as n increases without bound.
If an approaches a number, let’s call it L. Then it must be that L 
12
 2 L2  5L  12 and
2L  5
2 L2  5L  12  0   2 L  3 L  4   0  L  32 , 4 . Since the process won’t produce negative
numbers, the answer is
3
.
2
So the correct answer is A)
3
.
2
(October 2005, #19) If x 2  xy  15 x  12 and y 2  xy  15 y  42 , which of the following is a
possible value for x  y ?
Adding
 x  y
2
the
two
equations
leads
to
x 2  2 xy  y 2  15x  15 y  54
and
to
 15  x  y   54 . For z  x  y , you get z 2  15 z  54  0   z  18  z  3  0 . So
the possible values for x  y are 18,3 .
So the correct answer is A) 3.
(February 2006, #2) How many different four-digit numbers can be formed by arranging the
digits 2, 0, 0, and 6?
2
3
2
1
# of choices
for 1st digit
# of choices
for 2nd digit
# of choices
for 3rd digit
# of choices
for 4th digit
So it looks like there are 12 different four-digit numbers, but since two of the digits are zeros,
we have to divide by 2, giving us 6 of them.
So the correct answer is A) 6. [See the section on Sets and Counting]


k
(February 2006, #6) Let a, b  0 , M 
k
ln  an  
n 1
ln  bn  , N  e M , and P  k N . Then
n 1
P equals

k
M

k
ln  an  
n 1
Pk N k
ln  bn  
n 1
 ba 
k


k
ln  ba   k ln  ba  , N  eM  e
k ln
 ba 
  ba  ,
k
n 1
a
.
b
So the correct answer is A)
a
. [See the section on Logarithmic Properties]
b
(February 2006, #7) Which of the following imply that the real number x must be rational?
I. x5 , x7 are both rational
II. x6 , x8 are both rational
III. x5 , x8 are both rational
If x5 , x7 are both rational, then so is x 2 
rational, then consider x  2 .
 2
6
x7
x5
x3
3
and
,
and
so
is
x  2
x  2 . If x6 , x8 are both
5
x
x
x
 8 and
 2
8
 16 . If x5 , x8 are both rational, then so
x8
x5
x3
2
is x  5 and x  3 , and so is x  2 .
x
x
x
3
So the correct answer is B) I and III, only.
(February 2006, #8) A positive integer less than 1000 is chosen at random. What is the
probability it is a multiple of 3, but a multiple of neither 2 nor 9?
The 333 multiples of 3 are T  3,6,9,12,15, ,999 . The multiples of 2 in T would have to be
multiples of 6, and the 166 multiples of 6 in T is S  6,12,18, ,996 . The 111 multiples of 9
in T is R  9,18,17, ,999 . The numbers in T that are multiples of 2 and 9 would have to be
multiples of 18, and the 55 multiples of 18 in T is S R . The quantity of numbers less 1000
that are multiples of 3, but not multiples of 2 or 9 is 333  166  111  55  111 . So the
probability of selecting a positive integer less than 1000 that is a multiple of 3, but not a
111 1
multiple of 2 or 9 is
 .
999 9
So the correct answer is B)
1
.
9
(February 2006, #9) Let r and s be the solutions to the equation x2  3x  c  0 .
r 2  s 2  33 , then find the value of c.
The solutions of x2  3x  c  0 are
them together leads to
If
3  9  4c
3  9  4c
and
. Squaring and adding
2
2
36  8c
 9  2c . So we get that 9  2c  33  c  12 .
4
Or
We know that r  s  3  r 2  2rs  s 2  9  2rs  24  rs  12  c  12 .
So the correct answer is B) 12 . [See the section on Algebraic Formulas]
[See the section on Polynomial Properties]
(February 2006, #12) The midrange of a set of numbers is the average of the greatest and least
values in the set. For a set of six increasing nonnegative integers, the mean, the median, and the
midrange are all 5. How many such sets are there?
Call the six nonnegative integers x1  x2  x3  x4  x5  x6 .
We know that
x1  x6
 5,
2
x3  x4
x  x2  x3  x4  x5  x6
 5 , so we get that x1  x6  10 , x3  x4  10 , and
 5 , and 1
6
2
x2  x5  10 . Here are the possibilities:
x1
x2
x3
x4
x5
x6
1
0
1
2
8
9
10
2
0
1
3
7
9
10
3
0
1
4
6
9
10
4
0
2
3
7
8
10
5
0
2
4
6
8
10
6
0
3
4
6
7
10
7
1
2
3
7
8
9
8
1
2
4
6
8
9
9
1
3
4
6
7
9
10
2
3
4
6
7
8
So the correct answer is A) 10. [See the section on Statistics Formulas]
(February 2006, #13) The sum of the absolute values of all solutions of the equation
x3  4 x 2  6 x  22  x 2  2 x  2 can be written in the form a  b c , c a prime.
Find
a bc.
x3  4 x 2  6 x  22  x 2  2 x  2 leads to two equations:
x3  4 x 2  6 x  22  x 2  2 x  2  x 3  3x 2  8 x  24  0   x  3  x 2  8   0 with solutions
of  8, 3 .
x3  4 x 2  6 x  22    x 2  2 x  2   x 3  5 x 2  4 x  20  0   x  5   x 2  4   0
solutions of 2, 5 .
with
So the sum of the absolute values of all the solutions is
8  8  3  2  2  5  12  4 2 .
So the correct answer is E) 18.
(February 2006, #14) Find the number of three-digit numbers containing no even digits which
are divisible by 9.
So using only the digits 1, 3, 5, 7, and 9, we must make a three-digit number that is divisible by
9. In order for a number to be divisible by 9, the sum of its digits must be divisible by 9. The
three-digit numbers whose digit sum is 9 are arrangements of 1,1,7 and 1,3,5 and 3,3,3. The
three-digit numbers whose digit sum is 27 are arrangements of 9,9,9. So there are
3  6  1  1  11 .
So the correct answer is D) 11. [See the section on Divisibility Rules]
(February 2006, #15) If  is the acute angle formed by the lines with equations y  2 x  5 and
y  1  3x
y  1  3x , find tan  .
y  2x  5


     , so tan   tan      
tan   tan 
3  2

 1.
1  tan  tan  1   3 2 
So the correct answer is C) 1. [See the section on Trigonometric Formulas]
(February 2006, #16) Find the number of points of intersection of the unit circle and the graph
of the equation y 2  xy  x y  x x  0 .
For x  0 , the equation becomes y 2  xy  xy  x 2  0   y  x   0  y  x . So for x  0 ,
2
the graph looks like
For x  0 , the equation becomes y 2  xy  xy  x 2  0   y  x  y  x   0  y  x, y   x .
So for x  0 , the graph looks like
So putting all the pieces together along with the graph of the unit circle leads to
So the number of intersection points is 3.
So the correct answer is D) 3.
(February 2006, #17) Suppose that for a function y  f  x  , f  x   x for all x. Let A be the
point with x-coordinate a on the function y  f  x  and B be the point on the graph of the line
y  x for which AB is perpendicular to the line. Find an expression for the distance from A to
B.
A has coordinates  a, f  a   and B has coordinates  b, b  . The slope of the line through points
A and B is 1, so
f a  b
ab
 1  b 
a  f a
2
. Now apply the distance formula to get
2  f  a   a 
f a  a
 a  f a   f a  a 
.
D 

 
 
2
2
2
2

 

2
So the correct answer is C)
2
2  f  a   a 
2
. [See the section on Equations of Lines]
[See the section on Geometric Formulas]
(October 2006, #2) A fraction is chosen at random from all positive unreduced proper fractions
with denominators less than 6. Find the probability that the fraction’s decimal representation
terminates.
Here are the positive unreduced proper fractions with denominator less than 6:
1 1 2 1 2 3 1 2 3 4
, , , , , , , , , .
2 3 3 4 4 4 5 5 5 5
The only ones that don’t terminate are
So the correct answer is E)
1
2
8 4
and , so the probability is
 .
3
10 5
3
4
. [See the section on Probability Formulas]
5
(October 2006, #3) Two adjacent faces of a rectangular box have areas 36 and 63. If all three
dimensions are positive integers, find the ratio of the largest possible volume of the box to the
smallest possible volume.
36
63
The adjacent faces must share a side whose measurement is a common factor of 36 and 63.
These common factors are 1, 2, 3, and 9. The volume of the box is the product of 63 and this
common factor, so the largest possible volume is 63 times 9 and the smallest possible volume is
63 times 1. This means that the ratio of the largest volume to the smallest volume is 9.
So the correct answer is D) 9.
(October 2006, #5) In the expression  AM  AT YC  , each different letter is replaced by a
different digit 0 to 9 to form three two-digit numbers. If the product is to be as large as
possible, what are the last two digits of the product?
To get the largest product it must be  9M  9T 8C  , with C, T, and M replaced by 5, 6, and 7.
Here are the possibilities:
 97  96 85  791,520
 95 97 86   792,490
 96  9587   793,440
So the correct answer is B) 40.
(October 2006, #6) A basketball player has a constant probability of 80% of making a free
throw. Find the probability that her next successful free throw is the third or fourth one she
attempts.
Probability that the third free throw is successful: .2  .2  .8  .2  .8 .
2
Probability that the fourth free throw is successful: .2  .2  .2  .8  .2  .8 .
3
The probability that the third or fourth free throw is successful is
.2
2
.8  .2  .8  .2   .8  1.2  .0384 .
3
2
So the correct answer is B) .0384. [See the section on Probability Formulas]
 a b   5 10 0 0
(October 2006, #7) If 


 , find the smallest possible value of
 c d   3 6  0 0 
a  b  c  d , if a, b, c, and d are all positive integers.
Multiplying and equating the matrices leads to the system
5a  3b  0
10a  6b  0
5c  3d  0
.
10c  6d  0
This reduces into the system
5a  3b  0
5c  3d  0
.
So we get that
a  53 b
c  53 d
.
So a  b  c  d  85 b  85 d  85  b  d  . Since this must be a positive integer, b  d has to be a
positive multiple of 5. So the smallest possible value of a  b  c  d is 8.
So the correct answer is B) 8. [See the section on Matrix Multiplication]
(October 2006, #8) Sue works weekdays for $10 an hour, Saturdays for $15 an hour, and
Sundays for $20 an hour. If she worked 180 hours last month and earned $2315, how many
more weekday hours than Sunday hours did she work last month?
W  Sa  Su  180
10W  15Sa  20Su  2315
dividing the second equation by 5 leads to
Subtracting twice the first equation from the second equation leads to
Subtracting the second equation from the first equation leads to
W  Sa  Su  180
2W  3Sa  4Su  463
W  Sa  Su  180
Sa  2Su  103
W  Su  77
Sa  2Su  103
. So she worked
77 more weekday hours than Sunday hours.
So the correct answer is B) 77.
(October 2006, #10) The year 2006 is the product of exactly three distinct primes p, q, and r.
How many other years are also the product of three distinct primes with sum equal to
p  q  r?
2006  2  17  59 and p  q  r  78 . Let’s make adjustments to the prime factors to get new
prime factors which also sum to 78. To make 2 into a different prime factor means we have to
add an odd number to it, but adjusting the other prime factors by an odd number would make
them even and hence not prime, so the 2 must be left alone.
modifications:
2  17  6    59  6   2  23  53
2  17  12    59  12   2  29  47
Here are the allowable
.
So the correct answer is A) 2.
(October 2006, #11) How many positive integers less than 1000 are relatively prime to 105?
Two integers are relatively prime if their greatest common divisor is 1.
105  3  5  7 , so we have to count the number of positive integers less than 1000 that are not
multiples of 3, 5, or 7.
1000   # of multiples of 3   # of multiples of 5    # of multiples of 7 
  # of multiples of 15    # of multiples of 21   # of multiples of 35    # of multiples of 105 
 1000  333  200  142  66  47  28  9  457
So the correct answer is B) 457. [See the section on Sets and Counting]
(October 2006, #13) The equation xlog25 9  9log25 x  54 has a solution in common with which of
the following?
The solutions of A) are 1,125 .
The solutions of B) are 5 .
The solutions of C) are 5 .
The solutions of D) are 1, 5 .
The solutions of E) are 1, 5 .
125log25 9  9log 25 125   5  25 
log 25 9
9
log 25  525
 25log 25
9
 54
So the correct answer is A) x3  125 x 2  x  125  0 .
[See the section on Logarithmic Properties]
3
 25log 25 9  9log 25 5log 25 25  3  9  9 2  27  27
.
(October 2006, #19) Find the tens digit of 32007 .
Let’s find the pattern in the tens digits:
Power of 3
Tens digit
30
0
31
0
32
0
33
2
34
8
35
4
36
2
37
8
38
6
39
8
310
4
311
4
312
4
313
2
314
6
315
0
316
2
317
6
318
8
319
6
320
0
321
0
322
0
323
2
324
8
325
4
326
2
327
8
Starting with 30 , the tens digit repeats the same 20 numbers. From 30 to 32007 is 2008 tens
digits. 2008  20  100 r 8 , and the eighth number in the repeating 20 number sequence is 8.
So the correct answer is E) 8.
(October 2006, #20) In the sequence a1 , a2 , a3 ,
an1an2  2an an2  2an1an1 . Find
a2006
.
a2005
an1an2  2an an2  2an1an1  1  2 
sequence cn 
that cn  2 
, a1  1 , a2  2 , a3  5 , and for all n  3 ,
an
a
a
a
1
 2  n1  n  n1  .
an1
an2
an1 an2 2
Consider the new
an
1
5
for n  2 . We get that c2  2 , c3  , cn  cn1  , for n  3 . This means
an1
2
2
a
1
1
 n  2 for n  3 . 2006  c2006  2   2006  2   1004 .
a2005
2
2
So the correct answer is E) 1004.
(February 2007, #2) The teachers at Oak Tech have cars with average mileage 39,000 miles.
George buys a brand-new car, keeping his old car, and the average mileage drops to 36,400.
How many cars do the teachers now own?
Let T be the current total mileage of the cars and N be the current number of cars. So
T
T
 39,000 . We also have that
 36,400 , so 39,000 N  36,400  N  1 . This implies
N
N 1
36,400
that 2,600 N  36,400 , or N 
 14 .
2,600
So the correct answer is C) 14.
(February 2007, #3) The sequence log x,log x 2 ,log x3 ,log x 4 ,
is best described as which of
the following?
log x,log x 2 ,log x3 ,log x 4 ,
 log x,2log x,3log x,4log x,
 log x,log x  log x,log x  2log x,log x  3log x,
So the correct answer is C) arithmetic with common difference log x .
[See the section on Algebraic Formulas]
(February 2007, #4) A set of seven different positive integers has mean and median both equal
to 20. What is the largest possible value this set can contain?
Let’s call the seven positive integers x1  x2  x3  x4  x5  x6  x7 . The median being 20
implies that x4  20 , and the mean being 20 implies that the sum of the other values must be
120.
20
x1
x2
x3
x4
x5
x6
x7
We want the value of x7 to be as large as possible, so we need to make the values of
x1 , x2 , x3 , x5 , x6 as small as possible. This leads to
1
2
3
20
21
22
71
x1
x2
x3
x4
x5
x6
x7
So the correct answer is C) 71. [See the section on Statistics Formulas]
(February 2007, #7) If ln s  .6 and ln t  .9 , find log st e5.4 .
s  e.6 and t  e.9 , so st  e1.5 . log st e5.4  log e1.5 e5.4 
5.4
 3.6 .
1.5
So the correct answer is A) 3.6. [See the section on Logarithmic Properties]
(February 2007, #8) A function is symmetric to the origin and periodic with period 8. If
f  2   3 , what is the value of f  4   f  6  ?
A
particular
function
that
has
f  4   f  6   3sin   3sin 32  0  3  3 .
So the correct answer is B) 3 .
these
properties
is
f  x   3sin  4 x  .
(February 2007, #9) For how many integer values of k do the graphs of x  y  k and xy  k
NOT intersect?
Intersections correspond to real solutions of the system
x yk
xy  k
. Using substitution, you get
k
 k  x 2  kx  k  0 . In order for this quadratic equation to not have real solutions, its
x
discriminant must be negative. So we want k 2  4k  0 or k  k  4   0 . This leads to
x
k  1,2,3 .
So the correct answer is D) 3. [See the section on Algebraic Formulas]
(February 2007, #10) A point is chosen at random from the interior of a square of side 16.
Find the probability that the point is at least
2 units from both diagonals.
6
2
6
The point must be chosen from the four triangles outside of the central dashed cross. Each has
an area of 36 square units for a total of 144 square units. The area of the entire square is 256
144 9
square units, so the probability is
 .
256 16
So the correct answer is A)
9
. [See the section on Geometric Formulas]
16
[See the section on Probability Formulas]
(February 2007, #12) If cos  arctan  x    x (x in radians), then x 2 can be expressed in the form
a b
. Find a  b .
2
cos  arctan  x    x 
1
1  x2
 x   x2   x2  1  0  x2 
2
1  5
.
2
So the correct answer is A) 4. [See the section on Trigonometric Formulas]
x 2  3x  4
(February 2007, #17) If f  x  
, the inverse of f  x  can be written as
x 1
x 2  ax  b
1
. Find a  b  c .
f  x 
xc
Notice
f
1
 x
f  x   x  4; x  1,
that
so
f 1  x   x  4; x  5 .
This
means
that
x  4  x  5 x 2  9 x  20

.


x5
x5
So the correct answer is E) 34.
 x  y  kz  1

(February 2007, #18) Choose k so that the system  x  ky  z  2 is dependent. For which
kx  y  z  3

pair  x, y  below does there exist a z such that  x, y, z  will satisfy the resulting dependent
system?
x  y  kz  1
x  ky  z  2
subtract the first equation from the second equation, and subtract k times the
kx  y  z  3
x  y  kz  1
first equation from the third equation.  k  1 y  1  k  z  1
add the second equation to
1  k  y  1  k 2  z  3  k
x  y  kz  1
x  y  kz  1
the third equation.  k  1 y  1  k  z  1 simplify into  k  1 y   k  1 z  1 . For k  1, the
 2  k  k  z  2  k
2
 k  2  k  1 z  k  2
system has no solution, but for k  2 , you get the dependent system
solutions of y  z  13 , x  z  43 , z  z . The only pair that works is
z  43 .
So the correct answer is C)
 83 ,1 .
x  y  2z  1
3 y  3z  1
, which has
 83 ,1 , and it corresponds to
(February 2007, #19) A pentagon is circumscribed about a circle of diameter 6 cm. If the
pentagon has area 42 cm2 , find it perimeter in centimeters.
3
The area of the pentagon is
pentagon is 5 
s
15s
28
15s
, so we get that
 42  s  . So the perimeter of the
2
5
2
28
 28 .
5
So the correct answer is D) 28. [See the section on Geometric Formulas]
1
 1 
 4 
(February 2007, #20) The sum of the solutions of arctan    arctan 
  arctan 

 x
 x  2
 x  4
is
1
 1 
 4 
arctan    arctan 

arctan



x
 x  2
 x  4


1
 1 
 4 
tan arctan    arctan 
   tan arctan 
 
x
x

2
x

4









1
1
4
x 1
2
x  x2



 x2  5x  4  2 x2  4 x  2
2
1
1  x x  2  x  4
x  2x  1 x  4
 x 2  x  6  0  x  3, 2
But x  2 , so the sum of the solutions is 3.
So the correct answer is E) prime. [See the section on Trigonometric Formulas]
(February 2008, #3) The equation a3  b3  c3  2008 has a solution in which a, b, and c are
distinct even positive integers. Find a  b  c .
a  2n, b  2m, c  2l  8  n3  m3  l 3   2008  n3  m3  l 3  251 .
The cubic numbers we
need to consider are 1, 8, 27, 64, 125, 216, and the three that add up to 251 are 8, 27, and 216.
These yield n  2, m  3, l  6 , and this implies that a  b  c  2  2  3  6   22 .
So the correct answer is B) 22.
(February 2008, #4) For how many different integers b is the polynomial x2  bx  16
factorable over the integers?
The integer factor pairs of 16 are 1 and 16, -1 and -16, 2 and 8, -2 and -8, 4 and 4, -4 and -4.
These lead to values of b of 17, -17, 10, -10, 8, -8. So there are six possible values of b.
So the correct answer is E) 6. [See the section on Polynomial Properties]
(February 2008, #5) Let f  x   x 2  2 x  4 .
Which of the following is a factor of
f  x  f 2 y  ?
f  x   f  2 y   x 2  2 x  4 y 2  4 y   x 2  4 y 2    2 x  4 y    x  2 y  x  2 y   2  x  2 y 
  x  2 y  x  2 y  2 
So the correct answer is D) x  2 y  2 . [See the section on Algebraic Formulas]
(February 2008, #7) A fair coin is labeled A on one side and M on the other; a fair die has two
sides labeled T, two sides labeled Y, and two labeled C. The coin and die are each tossed three
times. Find the probability that the six letters can be arranged to spell AMATYC.
The outcome of the coin tosses must be AMA in any order. This has probability of
outcome of the die tosses must be TYC in any order. This has probability of
probability of being able to spell AMATYC is
So the correct answer is E)
2
. So the
9
3 2 1
  .
8 9 12
1
. [See the section on Probability Formulas]
12
(February 2008, #8) What is the value of  log624 625 log623 624 
 log6 7  log5 6  
 log6 7  log5 6 ?
log5 7
 log 5 6  log 5 7
log5 6
 log7 8 log6 7  log5 6    log7 8 log5 7  
log5 8
 log5 7  log5 8
log5 7
Continuing in this manner leads to
 log624 625 log623 624  log6 7  log5 6  log5 625  4 .
3
. The
8
So the correct answer is C) 4. [See the section on Logarithmic Properties]
(February 2008, #9) The letters AMATYC are written in order, one letter to a square of graph
paper, to fill 100 squares. If three squares are chosen at random, without replacement, find the
probability to the nearest 1/10 of percent of getting three A’s.
Writing the six letters in order will fill 96 squares, leaving room for an additional AMAT. So
the total number of A’s is 34, the total number of M’s is 17, the total number of T’s is 17, the
total number of Y’s is 16, and the total number of C’s is 16. So the probability of selecting
34  33  32
three A’s without replacement is
 .03700 .
100  99  98
So the correct answer is B) 3.7%. [See the section on Probability Formulas]
(February 2008, #10) A student committee must consist of two seniors and three juniors. Five
seniors are able to serve on the committee. What is the least number of junior volunteers
needed if the selectors want at least 600 different possible ways to pick the committee?
The
number
of
different
possible
ways
to
pick
the
committee
J  J  1 J  2 
J  J  1 J  2 
.
So
we
want
10 
 600
5 C2  J C3  10 
6
6
J  J  1 J  2   360 . So the least value of J is 9.
is
or
So the correct answer is D) 9. [See the section on Sets and Counting]
(February 2008, #12) Each bag to be loaded onto a plane weighs either 12, 18, or 22 pounds.
If the plane is carrying exactly 1000 lbs. of luggage, what is the largest number of bags it could
be carrying?
12 x  18 y  22 z  1000
x yz
or
6 x  9 y  11z  500 ,
500   3 y  5 z 
so
x
500  9 y  11z
,
6
and
. So to make x  y  z as large as possible, you need to have
6
500   3 y  5 z  divisible by 6 and 3 y  5 z as small as possible. Multiples of 6 less than 500
are 498, 492, 486, 480, 474, …. Choosing y  z  1 makes 500   3 y  5 z   492 , so the
largest number of bags is
492
 82 .
6
So the correct answer is C) 82.
(February 2008, #17) Let r, s, and t be nonnegative integers. For how many such triples
rs  t  24
is it true that r  s  t  25 ?
 r, s, t  satisfying the system 
r

st

24

Adding the equations together leads to
rs  t  r  st  48  r  s  1  t  s  1  48   r  t  s  1  48 . Here are the factor pairs of
48: 1 and 48, 2 and 24, 3 and 16, 4 and 12, 6 and 8. 1 and 48 produces no solutions. For 2 and
24, if s  1 , then we get 26 values of r and t that add up to 24. 3 and 16 produces no solutions.
4 and 12 produces no solutions. 6 and 8 produces no solutions.
So the correct answer is D) 26.
(October 2008, #3) If x  1 is one solution of ax2  bx  c  0 , what is the other solution?
c
If x  1 is one solution, then ax 2  bx  c   x  1 ax  c  . So the other solution is  .
a
c
So the correct answer is D)  . [See the section on Polynomial Properties]
a
(October 2008, #4) Ryan told Sam that he had 9 coins worth 45 cents. Sam said, “There is
more than one possibility. How many are pennies?” After Ryan answered truthfully, Sam said,
“Now I know what coins you have.” How many nickels did Ryan have?
Here are the possibilities:
# of pennies
# of nickels
# of dimes
# of quarters
Total value
5
0
4
0
45
5
3
0
1
45
0
9
0
0
45
So Ryan has 9 nickels.
So the correct answer is E) 9.
(October 2008, #5) A point  a, b  is a lattice point if both a and b are integers. It is called
visible if the line segment from  0,0  to  a, b  does not pass through any other lattice points.
Which of the following lattice points is visible?
b
b
. So if the fraction
can be reduced,
a
a
then the point  a, b  is not visible. Otherwise, it is visible. The only point in the list for which
The slope of the line segment from  0,0  to  a, b  is
b
can’t be reduced is  28,15 .
a
So the correct answer is B)  28,15 .
(October 2008, #6) A flea jumps clockwise around a clock starting at 12. The flea first jumps
one number to 1, then two numbers to 3, then three numbers to 6, then two to 8, then one to 9,
then two, then three, etc. What number does the flea land on at his 2008 th jump?
The sequence of jumps is 1, 2, 3, 2, 1, 2, 3,2 1, 2, 3, 2, …. If you break them into groups of
four, you get 8,8,8,8, …. Four goes into 2008 502 times, so the flea will have traveled a total of
502  8  4016 spaces on the 2008th jump. The position of the flea will be the remainder when
4016 is divided by 12: 4016  12  334 r8 .
So the correct answer is E) 8.
(October 2008, #8) All nonempty subsets of 2,4,5,7 are selected. How many different sums
do the elements of each of these subsets add up to?
Subset
Sum
2
4
5
7
2,4
2,5
2,7
4,5
4,7
5,7
2,4,5
2,4,7
2,5,7
4,5,7
2
4
5
7
6
7
9
9
11
12
11
13
14
16
2,4,5,7
18
There are 12 different sums.
So the correct answer is C) 12. [See the section on Sets and Counting]
(October 2008, #9) Luis solves the equation ax  b  c , and Ahn solves bx  c  a . If they get
the same correct answer for x, and a, b, and c are distinct and nonzero, what must be true?
ax  b  c  x 
cb
ac
cb ac
and bx  c  a  x 
, so

 bc  b2  a 2  ac
a
b
a
b
This rearranges into a 2  b 2  ac  bc  0   a  b  a  b   c  a  b   0  a  b  c  0 .
So the correct answer is A) a  b  c  0 . [See the section on Algebraic Formulas]
(October 2008, #12) The equation a6  b2  c 2  2009 has a solution in positive integers a, b,
and c in which exactly two of a, b, and c are powers of 2. Find a  b  c .
Since the power on a is the largest, let’s try to eliminate its term first. The largest value
possible for a is 3. With this value, we get b2  c 2  1280 . To find the other values as powers
of 2, let’s see how many factors of 2 are in 1280.
b 2  c 2  256  5  28  22  1  210  28   25    24  . So a  3, b  32, c  16 .
2
2
So the correct answer is E) 51.
(October 2008, #20) For all integers k  0 ,
P  k    22  21  1 22  21  1 24  22  1

22
k 1

 22  1  1 is always the product of two
k
integers n and n  1. Find the smallest value of k for which n   n  1  101000 .
First let’s generalize it into Q  k , x    x2  x  1 x2  x  1 x4  x2  1
x
2k 1
k
For k  0 , you get Q  0, x    x 2  x  1 x 2  x  1  1  x 4  x 2 .
For k  1, you get Q 1, x    x 2  x  1 x 2  x  1 x 4  x 2  1  1  x8  x 4 .
For k  2 , you get Q  2, x    x 2  x  1 x 2  x  1 x 4  x 2  1 x8  x 4  1  1  x16  x8 .
In general,

 x2  1  1 .
x
Q  k , x    x2  x  1 x 2  x  1 x 4  x 2  1
P  k   Q  k ,2   22
k 1
means that 22
1
k 2
 22
k 1
 22
k 1
 22
k 1

22
k 1
 101000 , so 22
k 1


 x2  1  1  x2
k
k 1
1 .
1
2k 1
So we want 22
k 2
k 1
 x2 .
Notice that

This

 22
k 1
 1  101000 .
k 1
 22 must be at least a 1,001 digit decimal number.
k 1
Let’s just force 22 1 to be a 1,001 digit decimal number, since 2n and 2n  1 always have the
same number of digits.
22
k 1
1
 101000   2k 1  1 log 2  1000  2k 1 
 1000 
1000
 1   k  1 log 2  log 
 1
log 2
 log 2 
 1000 
log 
1
log 2 

k
 1  10.697...
log 2
.
So the correct answer is C) 11.
(February 2009, #3) The perimeter of a rectangle is 36 ft and a diagonal is 170 ft. Its area in
ft 2 is
L2  W 2  170, L  W  18  L2  2LW  W 2  324  2LW  154  LW  77 .
So the correct answer is D) 77. [See the section on Algebraic Formulas]
(February 2009, #5) For what values of k will the equation x 14  7  kx2 have exactly two
real solutions?
The equation can be rearranged into kx2  x 14  7  0 . The discriminant of the quadratic
formula can be used to determine when there will be exactly two real solutions, for k  0 . For
this quadratic equation for k  0 , is 14  28k . In order to have exactly two real solutions,
1
14  28k  0  28k  14  k   . But this includes the value k  0 , where the equation is
2
not quadratic, and in fact has only one real solution.
1
So the correct answer k   , k  0 , which is not one of the possible answers.
2
[See the section on Algebraic Formulas]
(February 2009, #6) If x and n are positive integers with x  n and xn  xn1  xn2  2009 ,
find x  n .
x n  x n1  x n2  2009  x n2  x 2  x  1  2009  x n2  x 2  x  1  7 2  41 . So for x  7 and
n  4 , you get that x  n  11 .
So the correct answer is B) 11.
3
of the women are matched against half of the men.
7
What fraction of all the players is matched against someone of the other gender?
(February 2009, #7) In a tournament,
W  12 M
, and we know that 73 W  12 M , so plugging into the first expression
W M
3
3
6
6
7W  7W
7W
yields
 13  .
6
W  7W
13
7W
This fraction is
3
7
So the correct answer is D)
6
.
13
(February 2009, #11) At one point as Elena climbs a ladder, she finds that the number of rungs
above her is twice the number below her(not counting the rung she is on). After climbing 5
more rungs, she finds that the number of rungs above and below are equal. How many more
rungs must she climb to have the number below her be four times the number above her?
A  2B, A  5  B  5  2B  5  B  5  B  10, A  20 , so
4 15  S   15  S  5S  45  S  9
So the correct answer is E) 9.
(February 2009, #12) If sin  cos  .2 and sin 2  .96 , find sin 3   cos3  .
sin 3   cos3    sin   cos   sin 2   sin  cos  cos 2     sin   cos  1  12 sin 2 
 .2 1  .48   .296
.
So the correct answer is D) .296. [See the section on Trigonometric Formulas]
[See the section on Algebraic Formulas]
(February 2009, #13) How many asymptotes does the function g  x  
x
10 100 x  1
2
have?
It
lim
x 
has
vertical
x
10 100 x 2  1
asymptotes
1
 lim
x 
10 100 
1
x2
it has horizontal asymptotes of y 
x
of
1
10
and
x
1
x
and lim
 lim
x 
100
10 100 x 2  1 x

1
.
10
1
10 100 
Since
1
x2

1
,
100
1
1
and y  
. So it has four asymptotes.
100
100
So the correct answer is E) 4.
(February 2009, #14) For how many solutions of the equation x 2  4 x  6  y 2 are both x and y
integers?
x2  4 x  6  y 2  x2  4x  4  2  y 2   x  2  2  y 2  y 2   x  2  2
2
2
  y  x  2  y  x  2   2
.
Assuming x and y are integers and since 2 is prime There are only four possibilities:
y  x  2  1 y  x  2  2 y  x  2  1 y  x  2  2
,
,
,
y  x  2  2 y  x  2  1 y  x  2  2 y  x  2  1
None of these lead to integer solutions.
So the correct answer is A) 0.
(February 2009, #17) How many different ordered pairs of integers with y  0 are solutions
for the system of equations 6 x 2 y  y 3  10 xy  0 and 2 x 2 y  2 xy  0 ?
Adding
the
equations
8x 2 y  y 3  12 xy  0 .
yields
8x 2  y 2  12 x  0  y 2  8  x  34  
2
 4 x  3  0 , the
2
 4 x  3  7 which
2
Since
y  0,
we
get
that
9
2
 2 y 2   4 x  3  9 . Since y must be an integer and
2
only possible values for y are
won’t yield an integer solution.
1, 2 .
y  1 both lead to
y  2 both lead to  4 x  3  1 , and
2
hence to the ordered pair solutions of  1,2  and  1, 2  .
So the correct answer is B) 2.
(February 2009, #18) The graph of the equation x  y  x3  y 3 is the union of a
x  y  x3  y 3   x  y   x 2  xy  y 2    x  y   0   x  y   x 2  xy  y 2  1  0 .
So x  y  0 , which is a line, or x 2  xy  y 2  1  0 , which is a hyperbola.
So the correct answer is E) line and a hyperbola. [See the section on Algebraic Formulas]
(February 2009, #19) A four-digit number each of whose digits is 1, 5, or 9 is divisible by 37.
If the digits add up to 16, find the sum of the last two digits.
The possible four-digit numbers that meet the conditions are
the arrangements of three 5’s and one 1: 1555, 5155, 5515, 5551, none of which are divisible by
37
and the arrangements of two 1’s, one 9, and one 5: 1195,1159, 9115, 5119, 5911, 9511, 1591,
1951, 1519, 1915, 9151, 5191, of which only 1591 is divisible by 37.
So the correct answer is C) 10.
(October 2009, #1) Find the sum of the solutions to the equations x2  5x  6  0 and
x 2  4 x  3  0 which DO NOT satisfy both equations at once.
The solutions of x2  5x  6  0 are 6 and 1 from
 x  6 x  1  0 .
The solutions of
x 2  4 x  3  0 are 1 and 3 from  x  1 x  3  0 . So the sum of the solutions which don’t
satisfy both is 6  3  3 .
So the correct answer is E) 3.
(October 2009, #2) Four consecutive integers are substituted in every possible order for a, b, c,
and d. Find the difference between the maximum and minimum values of ab  cd .
For the four consecutive integers n, n  1, n  2, n  3 , the maximum value of ab  cd is
 n  3 n  2   n  n  1  2n 2  6n  6 ,
n  n  3   n  1 n  2   2n 2  6n  2 .
and
the
minimum
value
of
ab  cd
is
So the difference between the maximum and
minimum values of ab  cd is 4 .
So the correct answer is D) 4.
(October 2009, #3) The product of a number and b more than its reciprocal is y(b>0). Express
the number in terms of b and y.
If the number is x, then y  x  1x  b  , so y  1  bx , and x 
y 1
. So the answer is A.
b
So the correct answer is A)
y 1
.
b
(October 2009, #4) If f  x   x 2  x  2 , find the sum of all x values satisfying f  x  2   22 .
If f  x  2   22 , then x  2 must be a solution of y 2  y  2  22 . The solutions come from
y 2  y  20  0 or  y  5 y  4   0 . So x  7 and x  2 . So the sum of the x values is 5 .
So the correct answer is E) 5.
(October 2009, #5) Sue bikes 2.5 times as fast as Joe runs, and in 1 hr they cover a total of 42
miles. What is their combined distance if Sue bikes for .5 hr and Joe runs for 1.5 hr?
ratesue   2.5  rate joe , and
rate joe 
 rate
sue
 rate joe   1  42 , from which we can deduce that
42
42
42
42
and ratesue 
 2.5 . So the combined distance is
 2.5  .5 
 1.5  33 .
3.5
3.5
3.5
3.5
So the correct answer is C) 33.
(October 2009, #6) The equation a 4  b3  c 2  2009 (a, b, c positive integers) has a solution in
which a and b are both perfect squares. Find a  b  c .
b
1
a4
1
1
b3
1
4
256
4
64
9
729
a
2009  1  64   1944 ,
2009  1  1  2007 ,
2009  1  729   1279 ,
2009   256  1  1752 , 2009   256  64  1689 , 2009   256  729   1024 . Of the numbers
2007, 1944, 1279, 1752, 1689, and 1024, the only perfect square is 1024. So the numbers are
a  4, b  9, c  32 , and the sum a  b  c  45 .
So the correct answer is D) 45.
(October 2009, #7) How many 3-digit numbers have one digit equal to the average of the other
2?
Average digit
Other two digits
# of 3-digit numbers with these digits
1
0 and 2 or 1 and 1
5
0 and 4 or 1 and 3
2
11
or 2 and 2
0 and 6 or 1 and 5
3
or 2 and 4 or
17
3 and 3
0 and 8 or 1 and 7
4
or 2 and 6 or
23
3 and 5 or 4 and 4
1 and 9 or 2 and 8
5
or 3 and 7 or
25
4 and 6 or 5 and 5
3 and 9 or 4 and 8
6
or 5 and 7 or
19
6 and 6
5 and 9 or 6 and 8
7
13
or 7 and 7
8
7 and 9 or 8 and 8
7
9
9 and 9
1
The total is 121.
So the correct answer is E) 121.
(October 2009, #8) A rectangular solid has integer dimensions with length  width  height
and volume 60. How many such distinct solids are there?
60  22  3  5
Height
Width
Length
1
1
60
1
2
30
1
3
20
1
4
15
1
5
12
1
6
10
2
2
15
2
3
10
2
5
6
3
4
5
So there are 10 distinct rectangular solids.
So the correct answer is C) 10.
(October 2009, #9)
2sin x

cos x  sinx tan x
2sin x
2sin x
2sin x cos x
sin 2 x



 tan 2 x .
2
2
2
x
cos x  sinx tan x cos x  sin
cos
x

sin
x
cos
2
x
cos x
So the correct answer is A) tan 2x . [See the section on Trigonometric Formulas]
(October 2009, #10) If x 
1 3
1
 12 and y   , find the largest value of xy .
x 8
y
1 36
1 5
  xy 
 . Let
xy 8
xy 2
1 5
53
1
z  xy to get z   . Solving for z leads to 2 z 2  5 z  2  0 ,
 2, . So the largest
z 2
4
2
value is 2.
Multiplying the two equations together leads to
xy  1  1 
So the correct answer is D) 2.
(October 2009, #12) The sum of the squares of the three roots of P  x   2 x3  6 x 2  3x  5 is
 x  a  x  b  x  c   x3   a  b  c  x 2   ab  ac  bc  x  abc .
The
roots
of
2 x3  6 x 2  3x  5 are the same as the roots of x3  3x 2  32 x  52 . So it must be that the roots a,
b,
and
a  b  c
c
satisfy
the
a  b  c  3, ab  ac  bc  32 , abc   52 .
equations
 9  a 2  b 2  c 2  2ab  2ac  2bc  9  a 2  b 2  c 2  2 ab  ac  bc   9
2
 a 2  b2  c 2  2  32  9  a 2  b2  c 2  6
So the correct answer is B) 6. [See the section on Polynomial Properties]
(October 2009, #13) The value of 4
4
 1 1 1
log 2  2 4 28 216





 4log2
1
2 4 log
2
1
2 8 log
2
1
216 
 1 1 1
log 2  2 4 2 8 216





is
1
4
4
1 1 1
  
4 8 16
1 12
4
1
 42  2 .
So the correct answer is C) 2. [See the section on Algebraic Formulas]
(October 2009, #14) The figure shows a circle of radius 4 inscribed in a trapezoid whose longer
base is three times the radius of the circle. Find the area of the trapezoid.
y  m  x  12 
 x  4
2
  y  4   16
2
We need to find the value of m that will make the line tangent to the circle, so we want the
 x  4    y  4   16
system
to have just one solution.
This means
y  m  x  12 
2
2
2
 x  4   m  x  12   4  16  x 2  8x  16  m2  x  12   8m  x  12   16  16
2
2
 1  m2  x 2   24m2  8m  8 x  144m2  96m  16   0
that
4
must have a double root, so its discriminant must be zero. This implies that m   , and the
3
1
upper base measurement must be 6. So the area of the trapezoid is 2  6  12   8  72 .
So the correct answer is A) 72. [See the section on Geometric Formulas]
(October 2009, #16) The integer r  1 is both the common ratio of an integer geometric
sequence and the common difference of an integer arithmetic sequence. Summing the
corresponding terms of the sequences yields 7, 26, 90, … . The value of r is
b, br , br 2 ,
a, a  r , a  2r ,
Summing leads to a  b, a  r  br, a  2r  br 2 ,
, so we get the system
ab7
a  r  br  26
a  2r  br 2  90
Subtracting the first equation from the other two yields
r  br  b  19
2r  br 2  b  83
Subtracting twice the first equation from the second equation yields
br 2  2br  b  45  b  r  1  45  b  r  1  5  32 .
2
2
So r  1  3  r  4 .
So the correct answer is B) 4. [See the section on Algebraic Formulas]
(October 2009, #17) A hallway has 8 offices on one side and 5 offices on the other side. A
worker randomly starts in one office and randomly goes to a second and then a third office(all
different). Find the probability that the worker crosses the hallway at least once.
The probability of crossing the hallway at least once is equal to one minus the probability of not
crossing the hallway. The probability of not crossing the hallway is
8  7  6  5  4  3 396
3
3 10

 . So the probability that we want is 1   .
13  12  11
1716 13
13 13
So the correct answer is D)
10
. [See the section on Probability Formulas]
13
(October 2009, #19) In square ABCD, AB=10. The square is rotated 45 around point P, the
intersection of AC and BD . Find the area of the union of ABCD and the rotated square to the
nearest square inch.
5 2 5
C
D
P
B
A
So the area of the union is the area of square ABCD along with the area of the four small



triangles. This gives 100  4  12  2 5 2  5  5 2  5  100  100


2
2  1  117.157  117 .
So the correct answer is A) 117. [See the section on Geometric Formulas]
(October 2009, #20) The sum of the 100 consecutive perfect squares starting with a 2  a  0 
equals the sum of the next 99 consecutive perfect squares. Find a.
a 2   a  1   a  2  
2
2
  a  99    a  100    a  101   a  102  
2
2
2
  a  198 
2
2
Subtracting the left side from the right side leads to
 a  100 2  a 2    a  1012   a  12    a  102 2   a  2 2  

 
 

2
2
  a  198    a  98  


  a  99   0
2
This leads to
100  2a  100   100  2a  102   100  2a  104  
 100  2a  296    a  99   0 .
2
leads to
100 198a  100  102  104  106 
 296    a  99   0 . This leads to
100 198a  9900  2 1  2  3  4 
 98     a  99   0 . This leads to
2
2
100198a  9900  98  99   a  99   0 . So we get the equation
2
This
19800a  1960200  a 2  198a  9801  0  a 2  19602a  1950399  0 . The quadratic formula
yields
19602  196022  4  1950399 19602  19800

 19701 .
2
2
So the correct answer is a  19,701. [See the section on Algebraic Formulas]
(February 2010, #1) Let P  x   x3  2 x 2  3x  4 .
Find the largest prime factor of
P  4  P  2 .
P  4   64  32  12  4  40, P  2   8  8  6  4  2  P  4   P  2   38  19  2 .
So
the
largest prime factor of P  4   P  2  is 19.
So the correct answer is B) 19.
(February 2010, #2) A circle of radius 2 and center E is inscribed inside square ABCD. Find
the area that is inside ABE but outside the circle.
C
D
We want the area of the triangle minus the area of the quarter
circle: 12  2  4  14    22  4   .
E
So the correct answer is C) 4   .
2
[See the section on Geometric Formulas]
A
B
4
(February 2010, #3) The unique solution to the equation ax  b  10 is x  2 , and the unique
solution to the equation bx  a  8 is x  3 . Find a  b .
2a  b  10
3b  a  8
subtract twice the second equation from the first equation to get
5b  6
3b  a  8
now add
5b  6
3
times the first equation to the second equation to get
22 and divide the first equation
a
5
5
6
b
28
5
by 5 to get
. So a  b  .
22
5
a
5
So the correct answer is B)
28
.
5
be an arithmetic sequence with a0  2 , a3  a12  8 , and
(February 2010, #5) Let a0 , a1 ,
a5  0 . Find a3 .
2,2  d ,2  2d ,2  3d ,2  4d ,2  5d ,
So it must be that 2  3d   2  d   8 . This leads to d 2  d  6  0 and d  3,2 . Since
2
a5  0 , it must be that d  2 , so a3  2  3  2  8 .
So the correct answer is C) 8. [See the section on Algebraic Formulas]
(February 2010, #6) All solutions to the equation a3  b3  c 2  2010 (a, b, c positive integers)
have the same value for a  b . Find this value of a  b .
a 3  b3  c 2  2010   a  b   a 2  ab  b 2   2010  c 2 , so let’s check 2010  c 2 for factors of
11, 12, 13, 14, and 15:
11
12
13
14
15
44
2010  c 2
74
N
N
N
N
N
43
161
N
N
N
N
N
42
246
N
N
N
N
N
41
329
N
N
N
N
N
40
410
N
N
N
N
N
39
489
N
N
N
N
N
38
566
N
N
N
N
N
37
641
N
N
N
N
N
36
714
N
N
N
Y
N
35
785
N
N
N
N
N
34
854
N
N
N
Y
N
c
2010  362  714  14  51, so we need to find values of a and b so that a  b  14 and
a 2  ab  b2  51, but it’s not possible.
2010  342  854  14  61 , so we need to find values of a and b so that a  b  14 and
a 2  ab  b2  61, and a  9, b  5 and a  5, b  9 both work. So a  b  14 .
So the correct answer is D) 14.
(February 2010, #7) If z  a  bi (a and b real) and z 2  21  20i , a  b 
z 2   a  bi   a 2  2abi  b2   a 2  b2   2abi  21  20i . So a 2  b2  21 and ab  10 .
2
21  841
 10 
4
2
2
This leads to a  
 25 and b2  4 .
  21 and a  21a  100  0 . So a 
2
 a 
So a  b  5  2  7 .
2
2
So the correct answer is A) 7.
(February 2010, #8) A point C is chosen on the line segment AB such that
AC
BC
. Find

BC 5  AB
AC
.
BC
AC
BC
AC
1
AC
1


 AC  BC 

AC
BC 5  AB
BC 5  BC
BC 5   BC
 1
5 x 2  5 x  1  0 , and a solution of x 
So the correct answer is A)
or
x
1
.
5  x  1
This
5  25  20 5  3 5
.

10
10
5  3 5
. [See the section on Algebraic Formulas]
10

2010
(February 2010, #9) Let  x  represent the greatest integer  x . Find
log 5 n  .
n 1
1,2,3,4,5, ,24,25, ,124,125, ,624,625, ,2010
log5 0

log5 1
log5 2
log5 3
log5 4
2010
So
log 5 n   4  0  20  1  100  2  500  3  1386  4  7264 .
n 1
So the correct answer is E) 7264. [See the section on Logarithmic Properties]
leads
to
(February 2010, #10) If you roll three fair dice, what is the probability that the product of the
three numbers rolled is a prime?
The possible outcomes are 1,1,2 in any order, 1,1,3 in any order, and 1,1,5 in any order. There
9
1
are 9 outcomes with the product a prime, so the probability is
.

216 24
So the correct answer is B)
1
. [See the section on Probability Formulas]
24
(February 2010, #12) Three faces of a rectangular box that share a common vertex have areas
of 48, 50, and 54. Find the volume of the box.
If the three dimensions of the box are a, b, and c, then we have that ab  54 , ac  48 , and
bc  50 . The volume of the box is given by abc  ab  ac  bc  54  48  50  360 .
So the correct answer is A) 360.


(February 2010, #14) For a function f  x  , let f 2  x   f  f  x   , f 3  x   f f  f  x   , and
x2  1
on the domain  , 1
x2  1
so on. For the function f  x  
f
1
 x 
x2  1 2
, f  x 
x2  1
sequence is


x 2 1
x 2 1
x 1
x 2 1
2


2
2
x2  1
x2  1
,x, 2
,x,
x2  1
x 1
So the correct answer is B) x .
1
1

2 x2
x 2 1
2
x 2 1
 x,f
3
 x 
. So f 2010  x   x .
1,   ,
x 1
2
x 1
2

f 2010  x  
x2  1
,
x2  1
so
the
(February 2010, #16) A 100 m long railroad rail lies flat along level ground, fastened at both
ends. Heat causes the rail to expand by 1% and rise into a circular arc. To the nearest meter,
how far above the ground is the midpoint of the rail?
101
h
100
r
r
r
50.5
50

r h
So we get that
 r  h
2
 502  r 2  h2  2rh  2500  0 ,  
50.5
50.5
, and
 2   
2 r
r
2r  4r 2  10000
50
 r  r 2  2500 ,
sin   . From the quadratic formula, we get that h 
2
r
 50.5  50
so we just need to find the value of r. If we can solve the equation sin 
  , then we
 r  r
can plug it into the formula for h and get the result. An approximate solution is 206.8852258,
and
plugging
h  206.8852258 
it
 206.8852258
2
into
the
 2500  6.132899
.
formula
So the correct answer is E) 7. [See the section on Geometric Formulas]
[See the section on Algebraic Formulas]
yields
(October 2010, #1) A square is cut into two equal rectangles, each with a perimeter of 36.
Find the area of the square.
x
x
y
y
y
x
x
2 x  2 y  36
y  2x
So y  12 , and the area of the square is 144.
So the correct answer is D) 144.
(October 2010, #2) Last year, the cost of milk was 150% of the cost of bread. If the cost of
milk has risen by 20%, and the cost of bread has risen by 25%, what percentage of the current
cost of bread is the current cost of milk?
Milkthen  1.5Breadthen
Milknow  1.2Milkthen
Bread now  1.25Breadthen
So
Milknow  1.2 1.5Breadthen  
1.21.5 Bread
1.25
now
 1.44Bread now
So the correct answer is C) 144.
(October 2010, #3) Angles are complements if they add to 90 . Let A be nine times B ,
and the complement of B be nine times the complement of A . Find B .
m  A  9m  B 
90  m  B   9 90  m  A
So 90  m  B   9 90  9m  B  , which means that 80m  B   720 . So m  B   9 .
So the correct answer is C) 9 .
(October 2010, #4) Find the product of all values of x for which f  x  
x3
is
x  10 x  24
2
undefined.
f  x 
x3
x3

, so the product is 12  2  24 .
x  10 x  24  x  12  x  2 
2
So the correct answer is B) 24 .
(October 2010, #5) If you roll three fair dice, what is the probability that the product of the
three numbers rolled is even?
In order for the product not to be even, all three numbers would have to be odd. The probability
1 1 1 1
that all three numbers are odd is    . So the probability of rolling three numbers
2 2 2 8
1 7
whose product is even is 1   .
8 8
So the correct answer is E)
7
. [See the section on Probability Formulas]
8
(October 2010, #6) If f  x   ax 2  bx  c , f  1  10 , f  0   5 , f 1  4 , find f  2  .
a  b  c  10
c5
abc4
So
ab5
a  b  1
So a  2 , b  3 , and c  5 . Therefore f  2   2  4  3  2  5  7 .
So the correct answer is A) 7.
(October 2010, #7) A lattice point is a point with both coordinates integers. How many lattice
points lie on or inside the triangle with vertices  0,0  , 10,0  , and  0,8  ?
y  8  54 x
1
5
4
5
9
8
6
3
2
1
7
So there are 1  1  2  3  4  5  5  6  7  8  9  51 lattice points on or inside the triangle.
Or you could use Pick’s Theorem which says that the area of the triangle is equal to the number
of interior lattice points plus half the number of boundary lattice points minus 1. So you get
that 40  i  10  1. This means that the number of interior lattice points is 31. These 31
combined with the 20 boundary lattice points also give you a total of 51 lattice points.
So the correct answer is A) 51. [See the section on Geometric Formulas]
(October 2010, #8) The perimeter of a rectangle is 52, and its diagonal is 20. Find its area.
x
y
x  y  26
x 2  y 2  202
Squaring the first equation leads to
x 2  2 xy  y 2  262
x 2  y 2  202
Now subtract the second equation from the first to get
2 xy  262  202
xy 
 26  20  26  20   138
2
So the correct answer is B) 138. [See the section on Algebraic Formulas]
(October 2010, #9) The consecutive even numbers are written side-by-side to form an infinite
decimal 0.24681012141618 . Find the digit in the 2010th decimal place.
The one-digit evens(2,4,6,8) occupy positions 1 through 4.
The two-digit even numbers(10-98) occupy positions 5 through 94.
The three-digit even numbers(100-998) occupy positions 95 through 1444.
The four-digit even numbers(1000-9998) occupy positions 1445 through 19444.
To get from the 1444 position to the 2010 position requires 566 digits. This amounts to the first
141.5 four-digit even numbers. The 141st four-digit even number is 1280, and the 142nd fourdigit even number is 1282. So the digit in the 2010th place is a 2.
So the correct answer is A) 2.
(October 2010, #13) The equation a5  b2  c 2  2010 (a, b, and c positive integers) has a
solution in which b and c have a common factor d  1. Find d.
d 2  r 2  s 2   2009
The only possible values of a are 1, 2, 3, and 4. So we want to find d with
d 2  r 2  s 2   1978
d 2  r 2  s 2   1767
d 2  r 2  s 2   986
d 2  r 2  s 2   2009  49  41  7 2  52  42 
So d  7 .
So the correct answer is D) 7.
(October 2010, #17) The integer r  1 is both the common ratio of an integer geometric
sequence and the common difference of an integer arithmetic sequence. Summing the
corresponding terms of the sequences yields 7,26,90, . The value of r is
a,
ar ,
ar 2 ,
ar 3 ,
b, b  r , b  2r , b  3r ,
So
ab7
ar  b  r  26
ar 2  b  2r  90
Eliminating b leads to
ar 2  2ar  a  45
a  r 2  2r  1  45
a  r  1  45
2
a  r  1  5  32
2
So r  4 .
So the correct answer is B) 4. [See the section on Algebraic Formulas]
(October 2010, #19) Let P  x  be a polynomial with nonnegative integer coefficients. If
P  2   77 , and P  P  2   1874027 , then find the sum of its coefficients.
P  x   an x n  an1 x n1 
P  2   an 2n  an1 2n1 
 a1 x  a0
 2a1  a0  77
P  77   an 77n  an1 77n1 
 77a1  a0  1874027
So 77  a0 must be divisible by 2, and 1874027  a0 must be divisible by 77. This leads to
a0  1 . Now we have
an 2n1  an1 2n2 
an 77 n1  an1 77 n2 
 a1  38
 a1  24338
So 38  a1 must be divisible by 2, and 24338  a1 must be divisible by 77. This leads to a1  6 .
Now we have
an 2n2  an1 2n3 
 a2  16
an 77n2  an1 77n3 
 a2  316
So 16  a2 must be divisible by 2, and 316  a2 must be divisible by 77. This leads to a2  8 .
Now we have
an 2n3  an1 2n4 
 a3  4
an 77 n3  an1 77 n4 
 a3  4
So 4  a3 must be divisible by 2, and 4  a3 must be divisible by 77. This leads to a3  4 .
Now we have
an 2n6  an1 2n5 
an 77 n6  an1 77 n5 
 a4  0
 a4  0
So a4 , a5 , , an must all be zero.
So P  x   4 x3  8 x 2  6 x  1 , and the sum of its coefficients is 19.
So the correct answer is E) 19.
(October 2010, #20) If x  1  x  2 
 x  2010  m for all real numbers x, find the
maximum possible value for m.
a1 , a2 ,
minimum value at the median of a1 , a2 ,
f  x   x  a1  x  a2   x  an has
middle two values of a1 , a2 , , an  .
For an odd number of values
, an  , f  x   x  a1  x  a2 
, an  . For an even number of values a1 , a2 , , an  ,
its minimum value at any value at or between the
In this case, the minimum of f  x   x  1  x  2 
f 1005  1004  1003  1002 
So
 1  2  3 
 x  an has its
 x  2010 occurs for 1005  x  1006 .
 0 1  2  3 
 1005   1  2  3 
 1005
 1004   1005  2 
So the correct answer is B) 10052 . [See the section on Statistics Formulas]
1004  1005
 10052
2
Trigonometric Formulas:
1. sin 2 x  cos2 x  1
2. tan x 
sin x
cos x
3. cos 2 x  2cos 2 x  1
4. sin 2 x  2sin x cos x
5. tan  x  y  
tan x  tan y
1  tan x tan y
6. tan  x  y  
tan x  tan y
1  tan x tan y
a 2  b 2  c 2  2bc cos A
7. Law of Cosines: b 2  a 2  c 2  2ac cos B
c 2  a 2  b 2  2ab cos C
8. Law of Sines:
sin A sin B sin C


a
b
c
9. If y  x  180 , then cos x  cos y  0
Algebraic Formulas:
1. 1  2  3 
2. 12  22  32 
n
 n2 
n  n  1
2
n  2n  1 n  1
6
3.  a  b   a 2  2ab  b2
2
4.  a  b   a 2  2ab  b2
2
5. a 2  b 2   a  b  a  b 
6. a3  b3   a  b   a 2  ab  b 2 
7. a3  b3   a  b   a 2  ab  b 2 
8. Geometric Series:
a  ar  ar 2  ar 3 
 ar n1 
a  ar  ar 2  ar 3 
 ar n1 
a 1  r n 
a  ar n

first
last 1
1 r
1 r
a

; 1  r  1
1 r
9. Arithmetic Series:
 a  a   n  1 d 
 first

n  n  1
last
a   a  d    a  2d    a  3d     a   n  1 d   na  d 
 n

2
2




10. Quadratic Formula:
For the equation ax2  bx  c  0 with a, b, and c real numbers and a  0 , the solution(s) are
b  b 2  4ac
. If the discriminant, b2  4ac , is positive, then there are two real
2a
solutions; if it’s negative, then there are two imaginary solutions; and if it’s zero, then there is
one real solution.
given by
Logarithmic Properties:
1. logb xy  logb x  log b y; x, y  0
2. logb
x
 logb x  logb y; x, y  0
y
3. logb x r  r logb x; x  0
4. log a x 
logb x
logb a
5. blogb x  x; x  0
6. logb b x  x
Geometric Formulas:
1. Area of a triangle:
A  12  base  height
semiperimeter,
or Heron’s formula,
A  s  s  a  s  b  s  c  , where s is the
abc
.
2
2. Area of a parallelogram:
A  base  height
3. Area of a trapezoid:
A  12  base1  base2   height
4. Area of a circle:
A     radius 
2
5. Circumference of a circle:
C  2    radius    diameter
6. Distance between the points  x1 , y1  and  x2 , y2  :
d
 x2  x1 
2
  y2  y1 
2
7. Midpoint between the points  x1 , y1  and  x2 , y2  :
 x1  x2 y1  y2 
 2 , 2 


8. Angle sum of a triangle:
The sum of the angles in a triangle is 180 .
9. Pick’s Theorem:
In a square lattice, the area contained by a closed figure is equal to the number of interior lattice
points plus half the number of boundary lattice points minus 1.
Equations of Lines:
1. Point-Slope:
y  y1  m  x  x1 
2. Two-Point:
 y  y1 
y  y1   2
  x  x1 
x

x
 2
1 
3. Slope-Intercept:
y  mx  b
4. Intercept-Intercept:
x y
 1
a b
5. Slope:
m
y2  y1
x2  x1
Polynomial Properties:
1. Factor Theorem:
P  a   0 if and only if  x  a  is a factor of P  x  .
2. Remainder Theorem:
When P  x  is divided by  x  a  , the remainder is P  a  .
3. Division Theorem:
For
polynomials
D x
and
N  x ,
there
are
unique
polynomials
Q x
and
R  x  ( deg  R  x    deg  N  x   ) with D  x   N  x  Q  x   R  x  . Q  x  is called the quotient,
and R  x  is called the remainder.
4. Rational Zero Theorem:
If P  x   an x n  an1 x n1 
 a1 x  a0 is an nth degree polynomial with integer coefficients,
then the rational zeros of P  x  can be expressed in the form
p
, where p is a factor of a0 and q
q
is a factor of an .
5. Properties of Zeros:
The n zeros of the nth degree polynomial x n  an1 x n 
 a1 x  a0 , r1 , r2 ,
following:
r1  r2 
 rn   1 a0 , r1  r2 
n
 rn  an1 , …
, rn must satisfy the
Fundamental Theorem of Arithmetic:
Every positive integer greater than one can be written uniquely as a product of prime factors,
i.e. N  p1r1  p2r2   pkrk , where p1 , , pk are distinct prime numbers.
LCM and GCF:
For two whole numbers A and B with A  p1r1  p2r2 
LCM  A, B   p1
 p2
GCF  A, B   p1
 p2
max r1 ,s1 
min r1 ,s1 
 pkrk and B  p1s1  p s2 2 
max r2 ,s2 

 pk
min r2 , s2 

 pk
The formulas generalize to more than two numbers.
max rk ,sk 
min rk , sk 
 pksk ,
Sets and Counting:
For the universal set U
1. n  AC   n U   n  A 
2. n  A
3. n  A
B
B   n  A  n  B   n  A
C   n  A  n  B   n  C   n  A
4. The maximum possible value of n  A
B
B  n A C   nB
C   n A
B
C
B  is the minimum of n  A  and n  B  , and this
generalizes to any finite number of sets.
5. The minimum possible value of n  A
B  is n  A  n  B   n U  if this quantity is positive,
and zero otherwise.
6. The number of different subsets of a set with n elements, including the empty set, is 2n .
7. Fundamental Counting Principle:
If a decision process consists of k stages with the number of options equal to n1 , n2 ,
, nk ,
respectively, then the number of different ways of completing the decision process is
n1  n2   nk .
8. Combinations:
The number of different subsets with k elements from a set with n elements is
n!
C

.
n k
 n  k ! k !
9. Permutations:
The number of different arrangements in a line of k elements from a set with n
n!
elements(permutations) is n Pk 
.
 n  k !
The number of different arrangements in a line of n objects in which k1 of the objects are
identical, k 2 of the objects are identical,…, k j of the objects are identical, with
k1  k2 
 k j  n is
n!
.
k1 ! k2 !  k j !
Graph Properties:
1. Replacing x with  x in an equation in the two variables x and y has the effect of reflecting
the graph of the solutions of the equation across the y-axis.
2. Replacing y with  y in an equation in the two variables x and y has the effect of reflecting
the graph of the solutions of the equation across the x-axis.
Divisibility Rules:
1. A positive integer is divisible by 2 if and only if its one’s digit is even.
2. A positive integer is divisible by 3 if and only if the sum of its digits is divisible by 3. This
process may be repeated.
3. A positive integer is divisible by 4 if and only if the ten’s and one’s digits form a two-digit
integer divisible by 4.
4. A positive integer is divisible by 5 if and only if its one’s digit is either a 5 or a 0.
5. A positive integer is divisible by 6 if and only if it’s both divisible by 2 and divisible by 3.
6. A positive integer is divisible by 7 if and only if when you remove the one’s digit from the
integer and then subtract twice the one’s digit from the new integer, you get an integer
divisible by 7. This process may be repeated.
7. A positive integer is divisible by 7 if and only if when you remove the one’s digit from the
integer and then subtract nine times the one’s digit from the new integer, you get an integer
divisible by 7. This process may be repeated.
8. A positive integer with more than three digits is divisible by 7 if and only if when you split
the digits into groups of three starting from the right and alternately add and subtract these
three digit numbers you get a result which is divisible by 7.
9. A positive integer is divisible by 8 if and only if the hundred’s, ten’s, and one’s digits form a
three-digit integer divisible by 8.
10. A positive integer is divisible by 9 if and only if the sum of its digits is divisible by 9. This
process may be repeated.
11. A positive integer is divisible by 11 if and only if when you remove the one’s digit from the
integer and then subtract the one’s digit from the new integer, you get an integer divisible
by 11. This process may be repeated.
12. A positive integer is divisible by 11 if and only if when you subtract the sum of the ten’s
digit and every other digit to the left from the sum of the one’s digit and every other digit to
the left you get a number divisible by 11. This process may be repeated.
13. A positive integer with more than three digits is divisible by 11 if and only if when you split
the digits into groups of three starting from the right and alternately add and subtract these
three digit numbers you get a result which is divisible by 11.
Statistics Formulas:
1. Mean or average:
For the group of numbers x1 , x2 ,
, xn , the mean is x 
x1  x2 
n
 xn
.
2. Median:
For the ordered group of numbers x1  x2 
 xn , if n is odd, then the median is the middle
number, and if n is even, then the median is the average of the middle two numbers.
3. Mode:
The mode is the most frequently occurring number, if there is one.
Probability Formulas:
1. Equally-likely probabilities: If all the outcomes in the sample space, S, are equally-likely to
n E 
occur, then the probability that an event E occurs is given by P  E  
.
nS 
2. Complementary probabilities: P  E C   1  P  E  .
3.
Intersection probability: If the events E and F are independent, then
P  E F   P  E   P  F  . This formula generalizes for any finite number of independent
events.
4. Union probability: P  E
F   PE  PF   PE
5. Conditional probability: P  E | F  
PE
F
PF 
.
F .
6. Geometric probability: If a point is to be chosen at random from a region S, then the
area of E
probability that the point is in the region E is given by P  E  
.
area of S
Matrix Multiplication:
If the matrix A has n rows and k columns and the matrix B has k rows and m columns, then the
matrix product AB is defined and will have n rows and m columns. The entry in the ith row and
jth column of AB is the product of the ith row of A with the jth column of B. For example:
1 2
1 2 3
A
and B   2 3  , the product AB is defined, has 2 rows and 2 columns, and



4 5 6
 3 4 
1  1  2  2  3  3 1  2  2  3  3  4  14 20 
AB  
  32 47  .
4

1

5

2

6

3
4

2

5

3

6

4

 
