Download SMLE 2008

Interesting problems from the AMATYC Student Math League Exams 2008
(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)
3 2 1
  .
8 9 12
1
. [See the section on Probability Formulas]
12
3
. The
8
2
. So the
9
(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 .
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
C

C

10

10

 600
5 2 J 3
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
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
x yz
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
2
4
5
7
6
7
9
9
11
12
11
13
14
16
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 ,
2
P  k    22  21  1 22  21  1 24  22  1
2k 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
x2
k 1

 x2  1  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,

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
2
2k 1
 101000 , so 22
k 1


 x2  1  1  x2
k
k 1
1 .
1
k 1
x2
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 
log 
1
log 2 

k
 1  10.697...
log 2
So the correct answer is C) 11.
 1000 
1000
 1   k  1 log 2  log 
 1
log 2
log
2


.