Download SMLE 2006

Interesting problems from the AMATYC Student Math League Exams 2006
(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 
n 1

k
ln  an  
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
x  x2  x3  x4  x5  x6
x3  x4
 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 
   3 6   0 0  , find the smallest possible value of
c
d

 
 

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. Here are the allowable
modifications:
2  17  6    59  6   2  23  53
2  17  12    59  12   2  29  47
.
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
3
 25log 25 9  9log 25 5log 25 25  3  9  9 2  27  27
 54
So the correct answer is A) x3  125 x 2  x  125  0 .
[See the section on Logarithmic Properties]
(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
, a1  1 , a2  2 , a3  5 , and for all n  3 ,
an1an2  2an an2  2an1an1  1  2 
sequence cn 
that cn  2 
an
a
a
a
1
 2  n1  n  n1  .
an1
an2
an1 an2 2
Consider the new
an
5
1
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.