Download SMLE 2003

Interesting problems from the AMATYC Student Math League Exams 2003
(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.