Laney College
AMATYC Contest (Fall 2010)
SOLUTIONS
1. [D] The perimeter of each rectangle is three times the length of a side of the square.
2. [C] The cost of milk to that of bread was 150:100 last year, and so
(150)(120%):(100)(125%) this year, i.e. 180:125, or 36:25, i.e. 144:100.
3. [C] 90  B  9(90  9 B) . Solve it to get B  9 .
4. [B] The product of all roots of x 2  10  24  0 is the constant term  24 .
5. [E] The probability that the product is odd is 12  12  12  18 . So the answer is 1  18  78 .
6. [A] a  b  c  10 , c  5 , a  b  c  4 . The 3rd equation minus the 1st equation gives
2b  6 , so b  3 . The 3rd equation gives a  3  5  4 , so a  2 . So
f (2)  2(2) 2  3(2)  5  7 .
7. [A] The line segment joining (10, 0) and (0, 8) is of slope  54 , giving a lattice point
along that line segment for every horizontal distance of 5, i.e. 3 lattice points as a
whole. There are 11 9  99 lattice points on or inside the rectangle formed by
(0, 0) , (10, 0) , (10, 8) and (0, 8) . Of them 99  3  96 are off the diagonal
joining (10, 0) and (0, 8) , i.e. 48 on either side.. The answer is 48  3  51 .
8. [B]  2  w 2  20 2 ,   w  52 / 2 . But (  w) 2  ( 2  w 2 )  2w , so
26 2  20 2  2w . Thus w  138 .
9. [A] The number of 1-digit even numbers = 4. The number of 2-digit even numbers =
(the number of 1-digit numbers)  5= 9 5 . The number of 3-digit even numbers
= (the number of 2-digit numbers)  5= 90 5 . The even numbers so far
contribute 4 1  9  5  2  90  5  3  1444 digits. The remaining
2010  1444  566 digits come from 4-digit even numbers. 566  141 4  2 , so
the answer is the hundreds digit of the 142nd four-digit even number, 1282.
10. [A] A number is divisible by 7 precisely if, when the digits, starting from the ones
digit, are multiplied by 1, 3, 2,  1 ,  3 ,  2 , 1, 3, 2, …, respectively, the sum of
the products is divisible by 7. Since AMATYC is divisible by 7,
 2 A  3M  1A  2T  3Y  1C is divisible by 7. Let A be 1, M be 0. In order
for AMATYC to be divisible by 5, C has to be 5 (as 0 is taken by M.) So
2T  3Y  2 is divisible by 7. If T  2 , then Y  5 , which is taken by C already,
so it won’t work. If T  3 , then Y  2 . So 101325 is the smallest one possible.
11. [B] Want a number x such that f ( x)  ln 7 . I.e. ln( x  1  x 2 )  ln 7 , so
x  1  x 2  7 , so 1  x 2  7  x , 1  x 2  (7  x) 2 , so x 
13
12. [A] Let   arctan x , then x  tan  . But sin( 30   )  14
, so
13
sin 30 cos   cos 30 sin   14
, i.e. 12 cos 
cos to get
1
2

3
2
3
2
24
7
.
13
. Divide through by
sin   14
13
tan   14
sec , i.e. 7  7 3 tan   13 sec  . Square both
sides to get 49  98 3 tan   147 tan 2   169(1  tan 2  ) , so
22 tan 2   98 3 tan   120  0 . The quadratic formula gives tan  
5 3
11
5
.
13. [D] b  d , c  d . So a 5  d 2 ( 2   2 )  2010 , d 2 ( 2   2 )  2010  a .
a  1, 2, 3, 4 . For each of them, calculate 2010  a 5 and factor it completely to
1
Laney College
see if it has a double prime root. Only a  1 has such a property, with
2010  15  7 2  41 . So d  7 (with   4 ,   5 .)
14. [C] The problem boils down to calculating the percentage of the 1 x 1
square shown that is shaded. Note that AE is half of OD , so
BD  2 AB . But AB  CD . It follows that AB  BC  CD , so the
shaded area is 13 of the square.
15. [D] As shown in the accompanying picture, the area is
(4)(6)  2(2)(4)  2(2)(6)   (2) 2  76.566  77 .
A
E
B
C
D
O
16. [E] For f (x) to make sense, x has to satisfy x 2  1 . So, for f ( f ( x))
2
 2

to make sense, we see ( f ( x)) 2  1 , i.e.  x  1   1 , and so


x


x 2  1 , which is impossible.
1
x2
17. [B] The integer geometric sequence is a , ar , ar 2 , …. The integer arithmetic
ab  7


sequence is b , b  r , b  2r , ….. So  ar  b  r  26 . The 2nd minus the 1st
ar 2  b  2r  90

gives a(r  1)  r  19 , and the 3rd minus the 2nd gives ar (r  1)  r  64 .
Multiple a(r  1)  r  19 by r on both sides to get ar (r  1)  r 2  19r .
Comparing this with ar (r  1)  r  64 , we see that r 2  r  19r  64 . Solve this
quadratic equation to get r  4, 16 . Rule out 16, as a(r  1)  r  19 which would
read 15a  16  19 , would not give an integer for a .
18. [E] a1  p , a2  q , a3  pq , a 4  pq 2 , a5  p 2 q 3 , a 6  p 3 q 5 , a7  p 5 q 8 . So
p 5 q 8  12,500,000  2 558 , p  2 , q  5 . a8 / a 7  a 6  p 3 q 5  2 35 5  25000 .
19. [E] P( x)    a 4 x 4  a3 x 3  a 2 x 2  a1 x  a0 , where all coefficients are nonnegative
integers. It follows from the condition P(2)  77 that a1 , a 2 , etc are all strictly
less than 77. It also tells us that a0  77 . But if a0  77 , then P( x)  77 ,
contradicting P( P(2))  1874027 . So a0  77 too. Now P( P(2))  1874027
reads   a 4 77 4  a3 77 3  a 2 77 2  a1 77  a0  1874027 , making  a 4 a3 a 2 a1a0
a base-77 representation of 1874027. So repeatedly divide 1874027 by 77, the
resulting remainders give this representation  a 4 a3 a 2 a1a0 . This gives
a4  0, a3  4, a2  8, a1  6, a0  1 . So 4  8  6  1  19 is the answer.
20. [B] Note that | x  1 |  | x  2010 |  2009 , | x  2 |  | x  2009 |  2009  2  2007 ,
…, | x  1005 |  | x  1006 |  1006  1005  1 . It follows that
(1  2009)1005
| x  1 |    | x  2010 |  1  3  5    2009 
 1005 2 . The
2
equality holds when, for example, x is 1005.5.
2