Download SMLE 2011

Interesting problems from the AMATYC Student Math League Exams 2011
(February 2011, #1) After Ed eats 20% of a pie and Ahn eats 40% of a pie, Ed has twice as
much left as Ahn. Find Ed’s original amount of pie as a percentage of Ahn’s original amount.
Let E be Ed’s original amount of pie, and A be Ahn’s original amount of pie.
.8E  2 .6 A 
E 2  .6 3

  150% .
A
.8
2
So the correct answer is D) 150.
(February 2011, #2) The expression a # b  ab2  a for integers a, b  0 . If  a # b  #3  250 ,
find a  b .
 a # b  #3  250  9  ab2  a    ab2  a   250  ab 2  a  25  a b 2  1  25 .
The factor pairs of 25 are 1, 25 and 5, 5. This means that a  5 and b  2 .
So the correct answer is B) 7.
(February 2011, #3) Alicia always climbs steps 1, 2, or 4 at a time. For example, she climbs 4
steps by 1-1-1-1, 1-1-2, 1-2-1,2-1-1,2-2, or 4. In how many ways can she climb 10 steps?
Only 1’s
Ten 1’s
10!
1
10!
Only 2’s
Five 2’s
5!
1
5!
1’s and 2’s:
eight 1’s and one 2
six 1’s and two 2’s
Four 1’s and three 2’s
Two 1’s and four 2’s
9!
9
8!1!
8!
 28
6!2!
7!
 35
4!3!
6!
 15
2!4!
1’s and 4’s:
six 1’s and one 4
two 1’s and two 4’s
7!
7
6!1!
4!
6
2!2!
2’s and 4’s:
three 2’s and one 4
one 2 and two 4’s
4!
4
3!1!
3!
3
1!2!
1’s, 2’s, and 4’s:
four 1’s, one 2, and one 4
two 1’s, two 2’s, and one 4
6!
 30
4!1!1!
5!
 30
2!2!1!
This gives 1  1  9  28  35  15  7  6  4  3  30  30 .
OR
The number of different ways to get to the 5th step is equal to the number of different ways to
get to the 4th step plus the number of different ways to get to the 3rd step plus the number of
different ways to get to the 1st step. This is so because the 1st step plus 4 steps gets you to the
5th step, the 3rd step plus 2 steps gets you to the 5th step, and the 4th step plus 1 step gets you to
the 5th step. In general, an  an1  an2  an4 .
With the starting values of a1  1, a2  2, a3  3, a4  6 , let’s try this new scheme to get the rest.
a5  a4  a3  a1  6  3  1  10 , a6  a5  a4  a2  10  6  2  18 ,
a7  a6  a5  a3  18  10  3  31, a8  a7  a6  a4  31  18  6  55 ,
a9  a8  a7  a5  55  31  10  96, a10  a9  a8  a6  96  55  18  169
So the correct answer is E) 169. [See the section on Sets and Counting]
(February 2011, #4) The sum of six consecutive positive integers beginning at n is a perfect
cube. The smallest such n is 2. Find the sum of the next two smallest such n’s.
n  n  1  n  2  n  3  n  4  n  5  6n  15 .
Since this must be an odd number, we’ll only consider odd cubic numbers:
27, 125, 343, …. 27 gives the value 2, 6  2  15  27 . 729 gives the value 119.
From
the
list
of
answer
choices,
we
can
check
679  119  560, 680  119  561, 681  119  562, 682  119  563, 683  119  564 to see if
they generate a cubic number. 6  560  15  3375  153 .
So the correct answer is A) 679.
(February 2011, #5) The sum of the infinite geometric series S is 6, and the sum of the series
whose terms are the squares of the terms of S is 15. Find the sum of the infinite geometric
series with the same first term and opposite common ratio as S.
a  ar  ar  ar  ar 
2
3
4
a

, a2  a2r 2  a2r 4  a2r 6  a2r8 
1 r
a  ar  ar  ar  ar 
2
3
4
a


1 r
a2
1 r 2
a
1 r

15 5
 .
6 2
So the correct answer is B) 2.5. [See the section on Algebraic Formulas]
a2

1  r2
(February 2011, #11) Multiplying the corresponding terms of a geometric and an arithmetic
sequence yields 96, 180, 324, 567, …. Find the next term of the new sequence.
a, ar , ar 2 , ar 3 ,
b, b  d , b  2d , b  3d ,
, so we get
ab  96, ar  b  d   180, ar 2  b  2d   324, ar 3  b  3d   567 .
Assuming that a and b are
whole numbers, 96  25  3 , let’s try a  4 and b  24 .
This leads to the system
4r  24  d   180,4r 2  24  2d   324,4r 3  24  3d   567 , which has as a solution d  6 and
4,6,9, 272 , 814 ,
3
r  . These values lead to the sequences
2
24,30,36,42,48,
96,180,324,567,972, .
and the product sequence
So the correct answer is B) 972. [See the section on Algebraic Formulas]
(February 2011, #12) If log x y  log y x  2.9 and xy  128 , find x  y .
Assuming that x and y are whole numbers, then since 128  26 , the possible values for x and y
are x  1, y  128 , x  2, y  64 , x  4, y  32 , x  8, y  16 . For the pair x  4, y  32 , you get
log 4 32  log32 4 
5 2 29
 
 2.9 .
2 5 10
So the correct answer is B) 36.
(February 2011, #13) The equation a5  b2  c 2  2011(a, b, c positive integers) has a solution
in which two of the three numbers are prime. Find the value of the nonprime number.
The possible values of a are 1, 2, 3, and 4. For a  1, we get b2  c 2  2010 . For a  2 , we get
b2  c 2  1979 . For a  3 , we get b2  c 2  1768 . For a  4 , we get b2  c 2  987 . We can
eliminate a  1 and a  4 . So now we need to check a  1 and a  3 .
For a  1, 2010  382  566 , which is not a square, 2010  402  410 , which is not a square,
2010  422  246 , which is not a square, 2007  442  71 , which is not a square,
2007  462  109 .
For a  3 , 1768  382  324 , which is 182 , but 18 is not a prime. 1768  402  402 , which is
not a square, 1768  422  4 , which is 22 , so we get 3, 2, and 42.
So the correct answer is C) 42.
(February 2011, #14) A palindrome is a number like 121 or 1551 which reads the same from
right to left and from left to right. How many 4-digit palindromes are divisible by 17?
4-digit palindromes are of the form abba, where a is 1,2,3,4,5,6,7,8,or 9 and b is
0,1,2,3,4,5,6,7,8,9. Now abba  1000a  100b  10b  a  1001a  110b  11 91a  10b  , so we
can just examine the 4-digit numbers which are multiples of both 17 and 11, and hence just
multiples of 187. We can skip multiples of 10.
187  6
1122
187  19
3553
187  31
5797
187  42
7854
187  7
1309
187  21
3927
187  32
5984
187  43
8041
187  8
1496
187  22
4114
187  33
6171
187  44
8228
187  9
1683
187  23
4301
187  34
6358
187  45
8415
187  11
2057
187  24
4488
187  35
6545
187  46
8602
187  12
2244
187  25
4675
187  36
6732
187  47
8789
187  13
2431
187  26
4862
187  37
6919
187  48
8976
187  14
2618
187  27
5049
187  38
7106
187  49
9163
187  15
2805
187  28
5236
187  39
7293
187  51
9537
187  17
3179
187  29
5423
187  41
7667
187  52
9724
And 187  53  9911 .
So the correct answer is B) 4.
(February 2011, #16) The increasing sequence of positive integers a1 , a2 , a3 ,
satisfies the
equation an2  an  an1 for all n  1 . If a7  160 , find a8 .
The sequence is a1 , a2 , a1  a2 , a1  2a2 ,2a1  3a2 ,3a1  5a2 ,5a1  8a2 ,8a1  13a2 ,
. We know
that 5a1  8a2  160 , which means that a2 must be a multiple of 5. If we go with a2  5 , then
a1  14 , but it doesn’t work.
a8  8a1  13a2  8  16  13  10  258 .
So the correct answer is B) 258.
If we go with
a2  10 , then
a1  16 , it works.
(October 2011, #1) If the standard order of operations is reversed (that is, additions and
subtractions are done first and exponentiation is done last), what is the value of 2  3 ^ 2  3 ?
23^ 2  3
23^ 5
6^5
7776
So the correct answer is E) 7776.
(October 2011, #2) The price of a stock rose 20% on Monday, fell 10% on Tuesday, and
1
increased by
on Wednesday. By what percent did the price rise from Monday to
6
Wednesday?
1.2  S  .9 
7
 1.26S
6
So the correct answer is B) 26.
(October 2011, #3) The system of equations ax  by  8 and ax  by  20 has the solution
 x, y    2,3 .
Find a  b .
Plugging in  2,3 leads to
2a  3b  8
2a  3b  20
get that b  2 .
So the correct answer is D) 9.
. Adding the two equations leads to a  7 , so we also
(October 2011, #4) The positive integers a, b, and c satisfy a6  b2  c 2  2011 .
a bc.
Find
The possible values of a are 1, 2, and 3. These lead to b2  c 2  2010 with b  c  49 to 53 ,
b2  c 2  1947 with b  c  48 to 52 , and b2  c 2  1282 with b  c  47 to 51
With a  3 , you get
b
1
c  1282  b 2
35.79106
2
35.74913
3
35.67913
4
35.58089
5
35.4542
6
35.29873
7
35.1141
8
34.89986
9
34.65545
10
34.38023
11
34.07345
12
33.73426
13
33.36165
14
32.95451
15
32.51154
16
32.03123
17
31.5119
18
30.95158
19
30.34798
20
29.69848
21
29
So the correct answer is D) 53.
The previous table can be generated on a TI-83 calculator by pressing the ‘Y =’ key, entering
1282  X  , pressing the ‘TABLE’ key, and scrolling down.
2
(October 2011, #5) Different shades of pink, red, and white can be made by mixing whole
numbers of quarts of red and white paint. Shades are different if the ration of red to white paint
is different. Find the number of different possible shades that can be made from at most 4
quarts of red and 5 quarts of white paint.
The total number of ratios is 4  5  20 , but some of them are equivalent. 1:1, 2:2, 3:3, and 4:4
are equivalent. 1:2 and 2:4 are equivalent. 2:1 and 4:2 are equivalent. This gets us down to
only 15 different ratios.
So the correct answer is A) 15.
[See the section on Sets and Counting]
(October 2011, #6) The function y  f  x  has zeros 2 and 6.
Find the zeros of
y  3 f  2  2 x  .
2  2 x  2  x  2 and 2  2 x  6  x  2
So the correct answer is A) 2, 2 .
(October 2011, #7) One population P1  t  grows exponentially at the same rate that another
population P2  t  decays exponentially. If the populations were both equal to P on Jan. 1 2009,
how will the populations be related on Jan. 1 2012?
P1  t   Pe at and P2  t   Pe  at , so P1  t  P2  t   P 2 .
So the correct answer is B) P1  t  P2  t   P 2 .
(October 2011, #8) For b  c  0 , both x2  bx  8 and x 2  cx  8 factor over the integers.
Find b  c .
It must be that x 2  bx  8   x  8  x  1  x 2  9 x  8 and
b
x  cx  8   x  2  x  4   x  6 x  8
2
2
c
So the correct answer is C) 3.
(October 2011, #9) Ed drives from San Mateo to Atascadero, a distance of 197.5 mi. He starts
driving at a constant speed and reduces his speed by 5 mph after each half hour of driving. If
the trip takes 3 hr 20 min, how far did he travel in the first 2 hours?
1
2
 R  12   R  5   12   R  10   12   R  15   12   R  20   12   R  25   13   R  30   197.5
10
3
R  952  197.5
20 R  285  1185
R  73.5
1
2
 73.5  12   73.5  5  12   73.5  10   12   73.5  15  132
So the correct answer is B) 132.
(October 2011, #10) Sun fills her 10 liter radiator with 20% antifreeze and 80% water. She
removes some of the mixture and replaces it with antifreeze. If the radiator is now one quarter
antifreeze, how many liters of the original mixture did she remove?
2  .2 x  x 1

10
4
.8 x  2 1

10
4
3.2 x  8  10
3.2 x  2
x  .625
So the correct answer is D) .625.
(October 2011, #11) How many numbers with no more than six digits can be formed using
only the digits 1 through 7, with no digit used more than once in a given number?
One-digit numbers: 7
Two-digit numbers: 7  6  42
Three-digit numbers: 7  6  5  210
Four-digit numbers: 7  6  5  4  840
Five-digit numbers: 7  6  5  6  3  2520
Six-digit numbers: 7  6  5  4  3  2  5040
So the correct answer is E) 8659.
[See the section on Sets and Counting]
(October 2011, #12) The lines with equations 2 x  3 y  24 and 3x  2 y  6 are symmetric
with respect to a line with equation y  mx  b with m  0 . Find m  b .
3x  2 y  6
2 x  3 y  24
The line that we want must bisect the obtuse angle between the two given lines. The acute
angle between the lines can be determined from the formula
 23    32 
m1  m2
5
1 5
tan  


,
so
the
obtuse
angle
would
be
. We need
180


tan
1  m1m2 1    23   23  12
12
half of this angle for the bisector, and this would be 90  12 tan 1
5
. And we need to add this
12
5
 2
 2
angle to tan 1    , giving us 90  12 tan 1  tan 1    . The slope of the line we want is
12
 3
 3

5
 2 
the tangent of this angle , tan  90  12 tan 1  tan 1     . From some trig identities, we get
12
 3 


1
5
1 cos tan

2
12 
 1  2  


2

1 5 
1 5 
1
1

tan  90  2 tan

tan
tan

cot
tan

5






2

sin  tan 1 
3
12 
12  3

 3 



  12 1 5   1 . The line
1 cos tan
5 


 2 
1 5 

2
1  tan  90  12 tan 1  tan  tan 1     1  23 cot  12 tan
 1  3   1 512 
12 
12  


 3 
sin  tan

12


passes through the point of intersection of the given lines which is  6,12  , so an equation for
the line is y  12  x  6  y  x  18 .
So the correct answer is D) 19.
[See the section on Trigonometric Formulas]
OR
Lines with reciprocal slopes are bisected by lines with slope of 1 and 1:
yx
y  mx
y  x
1
1
m
y
1
x
m
1
m
1
The lines with equations 2 x  3 y  24 and 3x  2 y  6 have reciprocal slopes, so the bisecting
line with positive slope must have slope of 1.
[See the section on Equations of Lines]
(October 2011, #13) A square of area 45 is inscribed in a circle C. Find the area of a square
inscribed in a semicircle of circle C. (Inscribed means having all 4 vertices on a given figure.)
From November 2003, #13: Square ABCD is inscribed in circle O, and its area is a. Square
EFGH is inscribed in a semicircle of O. What is the area of square EFGH?
H
G
D
C
E
F
A
B
The area of an inscribed square in a circle of radius r, is 2r 2 .
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
2a
.
5
Using this, we get that the area of the square inscribed inside the semicircle is
2  45
 18 .
5
So the correct answer is B) 18.
(October 2011, #14) The left edge of a dollar bill is folded against the bottom edge to form an
isosceles right triangle at the left edge. The new left edge is again folded against the bottom
edge. A vertex of the new triangle is the upper right corner of the bill. If a dollar bill is 157
mm long, find its width to the nearest millimeter.
157  W
W
2W
157  W
W
157  W 
2W
From the labeled diagram, it must be that
2W W  157



2  1   0 . So W  157

So the correct answer is C) 65.

2W 

157  W 
2  1  65.031... .
2
W
2
W 2
 W 2  157 . This simplifies into
(October 2011, #15) Five boxes are placed inside an empty box. Each of the 5 new boxes is
either left empty or has 5 new boxes placed inside it. This process is repeated until there are 18
boxes containing other boxes. Find the number of empty boxes.
Here are 18 boxes containing other boxes, and there are 73 empty boxes.
So the correct answer is A) 73.
(October 2011, #16) Al, Bo, Cy, and Di are to receive math, physics, chem, and bio awards. Al
thinks Di will win bio, Bo thinks Cy will win chem., Cy thinks Al won’t win mth, and Di thinks
Bo will win physics. The math and bio winners are both right, and the other winners are both
wrong. Who wins the math award?
Let’s try to make 2 of them right and 2 of them wrong:
Al and Bo are right, Cy and Di are wrong.
Al
Bo
Cy
Di
chem bio
Not possible.
Al and Cy are right, Bo and Di are wrong.
Al
Bo
Cy
Di
chem math phys bio
Al
Bo
Cy
Di
phys chem math bio
Possible, but the math and bio winners aren’t both right.
Al and Di are right, Bo and Cy are wrong.
Al
Bo
phys
Not possible.
Cy
Di
bio
Bo and Cy are right, Al and Di are wrong.
Al
Bo
Cy
Di
bio math chem phys
Al
Bo
phys bio
Cy
Di
chem math
Possible, but the math and bio winners aren’t both right.
Bo and Di are right, Al and Cy are wrong.
Al
Bo
Cy
Di
bio phys chem math
Possible, but the math and bio winners aren’t both right.
Cy and Di are right, Al and Bo are wrong.
Al
Bo
Cy
chem phys bio
This is it.
So the correct answer is D) Di.
Di
math
(October 2011, #17) The digits 1 through 9 are separated into 3 groups of three digits, and the
product of each group is found. Let P be the largest of the 3 products. Find the smallest
possible value of P.
74 is not the product of 3 digits 1 through 9. 73 is not the product of 3 digits 1 through 9. 71 is
not the product of 3 digits 1 through 9. So it’s down to either 70 or 72. 70 doesn’t work:
70  2  5  7 , and the other groups would consist of 9, ?, ? and 8, ? , ?, but no matter how you
position the 1, 3, 4, and 6, you will get a product that’s larger than 70. With 72, you can get the
groups 1, 8, 9 and 6, 3, 4 and 7, 2, 5 and the largest product is 72.
So the correct answer is C) 72.
(October 2011, #18) Out of 10 red chips and 15 green chips, 6 are placed into a bag, 10 into a
box, and 9 into a bowl. In how many ways can the chips be distributed, if only the number of
red and green chips in each container matters?
Let r1 be the number of red chips in the bag, r2 be the number of red chips in the box, r3 be the
number of red chips in the bowl. Let g1 be the number of green chips in the bag, g 2 be the
number of green chips in the box, g3 be the number of green chips in the bowl. This leads to
the system of equations
r1  r2  r3  10
g1  g 2  g3  15
r1  g1  6
r2  g 2  10
r3  g 3  9
Which has solutions given by g3  t , g 2  s, g1  15  s  t , r3  9  t , r2  10  s, r1  t  s  9 ,
where s and t are nonnegative integers with 0  s  t  9 . The number of different values of red
chips and green chips in each container is equal to the number of different pairs of values of s
and t.
s
t
This is equal to 10  9  8  7  6  5  4  3  2  1 
So the correct answer is D) 55.
10  11
 55 .
2
(October 2011, #19) Square ABCD has side length 72. Let E be the midpoint of side AB , and
let BD and CE intersect at G. Find the length of the altitude to BE in GEB .
72
B
C
36
x
G
72 - x
72
E
36
A
72
The area of the square can be calculated as
area  CBE   area  DAB   area  CGD   area  BGE  .
This leads to the equation 18  72  36  72  36  72  x   18 x  722 .
This simplifies into 18  72  54 x  x  24 .
So the correct answer is D) 24.
D
(October 2011, #20) Let r be the positive real zero of P  x   9 x5  7 x 2  9 .
r 4  2r 9 
 kr 5k 1 
can be expressed as the rational number
a
in lowest terms. Find
b
a  b.
Recall the geometric series formula:
1  x  x 2  x3 
 x n1 

1
; x  1.
1 x
Notice that
1  2 x  3x 2  4 x3 
 nx n1 

1  x  x 2  x3  x 4 

1
1 x

x
1 x

x  x 2  x3  x 4 

x x x 
2
3
4
x2

1 x

x x 
3

4
x4 
The sum
x3

1 x
x4

1 x
1
x
x2
x n1



 

1 x 1 x 1 x
1 x
1

1  x  x 2  x3  x 4  

1 x
1

; x 1
2
1  x 
r 4  2r 9 
 r 4 1  2r 5  3r10  4r15 

r4

2
1  r 5 
 kr 5 k 1 
 k r5 
k 1



But r is the real zero of P  x   9 x5  7 x 2  9 , so 9r 5  7r 2  9  0  r 5  1  79 r 2 . Plugging
this into the previous equation leads to
r  2r 
4
9
 kr
5 k 1



r4
7
9
r2 
So the correct answer is E) 130.
2

81
.
49
[See the section on Algebraic Formulas]