# 02 AMC10 Solutions

The MATHEMATICAL ASSOCIATION OF AMERICA American Mathematics Competitions 3rd Annual American Mathematics Contest 10

AMC 10 - Contest P
Solutions Pamphlet
January 2002
The “P” set of contests were originally developed for a group of schools in Taiwan, to be taken in early January of 2002. Once the contests were taken, the AMC office released the questions here for use as a practice (“P”) set of questions for the 2002 AMC 10 and AMC 12. This Pamphlet gives at least one solution for each problem on this year’s contest and shows that all problems can be solved without the use of a calculator. When more than one solution is provided, this is done to illustrate a significant contrast in methods, e.g., algebraic vs geometric, computational vs conceptual, elementary vs advanced. These solutions are by no means the only ones possible, nor are they superior to others the reader may devise. We hope that teachers will inform their students about these solutions, both as illustrations of the kinds of ingenuity needed to solve nonroutine problems and as examples of good mathematical exposition. However, the publication, reproduction, or communication of the problems or solutions of the AMC 10 during the period when students are eligible to participate seriously jeopardizes the integrity of the results. Duplication at any time via copier, phone, email, the Web or media of any type is a violation of the copyright law.
Orders for prior year Exam questions and Solutions Pamphlets should be addressed to:

American Mathematics Competitions University of Nebraska, P.O. Box 81606 Lincoln, NE 68501-1606

Copyright ? 2002, Committee on the American Mathematics Competitions The Mathematical Association of America

Solutions 2002 AMC 10 P 1. (C) Since (24 )8 = 232 and (48 )2 = 416 = 232 , we have (24 )8 232 = = 1. (48 )2 232 2. (B) Let n be the smallest of the eleven integers. Then 2002 = n + (n + 1) + (n + 2) + . . . + (n + 10) = 11n + (1 + 2 + 3 + . . . + 10) = 11n + (10)(11)/2 = 11n + 55. Solving 11n + 55 = 2002 yields n = 177.

2

3. (D) Imagine six spaces to hold the six digits. We can place the two 1s in two of 6! the spaces in 6 2 = 2!4! = 15 ways. (The digits 2, 0, 0, and 2 can then occupy the remaining four spaces.) 4. (C) For 4x 5y 6z = 22x+z 3z 5y both to be a perfect square, the exponent on each prime must be even. That is, y and z must be even. Only choice (C) satis?es this condition. 5. (C) Since an+1 = an + 1 3

2 3 3 = an?2 + 3 . . . n = a1 + 3 = an?1 + We see that a2002 = 1 +
2001 3

= 668.

6. (D) Let the sides of the rectangle be l and w. We are given that l + w = 50 and that l2 + w2 = x2 . Now observe that 2lw = (l + w)2 ? (l2 + w2 ) = 2500 ? x2 and hence the area of the rectangle is lw =
2500?x2 2

= 1250 ?

x2 2 .

Solutions 2002 AMC 10 P

3

7. (B) Let a, b, c be the dimensions of the box, a ≤ b ≤ c. Since abc = 2002 = 2·7·11·13, the only possible triples (a, b, c) are (1,1,2002), (1,2,1001), (1,7,286), (1,11,182), (1,13,154), (1,14,143), (1,22,91), (1,26,77), (2,7,143), (2,11,91), (2,13,77), (7,11,26), (7,13,22), and (11,13,14). Among these, the last triple gives the minimum sum, 38. 8. (E) Because 64 = 26 , we see that 64 is a sixth power, a cube, a square, as well as a ?rst power. Therefore, z = 6, 3, 2, or 1. For z = 6 we have only (x, y ) = (2, 1); for z = 3 we have (x, y ) = (4, 1), and (2, 2); for z = 2 we have (x, y ) = (8, 1) and (2, 3); and for z = 1 we have (x, y ) = (64, 1), (8, 2), and (2, 6). There are nine solutions in all. 9. (B) Checking the ?rst few values, we ?nd u0 u1 u2 u3 u4 =4 = f (4) = 5 = f (5) = 2 = f (2) = 1 = f (1) = 4

u5 = f (4) = 5 u6 = f (5) = 2. In general, we see that u4k+j = uj , where k is any integer greater than or equal to zero. Hence, u2002 = u4·500+2 = u2 = 2. 10. (C) Multiplying both sides of the given equation by b(b + 10a) yields 2ab + 10a2 + 10b2 = 2b2 + 20ab which is equivalent to (a ? b)(5a ? 4b) = 0. Since a = b, we have 5a = 4b, or a b = 0.8. 11. (A) The expression x?2 is a factor of P (x) if and only if P (2) = 0 = 8k +8k 2 +k 3 = √ 32 k (k 2 + 8k + 8). The values of k are therefore 0 and ?8± and the sum of 2 these real numbers is ?8. 12. (C) Evaluating the four values we ?nd that (f11 (a)f13 (a))14 = (a24 )14 = a336 f11 (a)f13 (a)f14 (a) = a38 (f11 (f13 (a)))14 = (a13·11 )14 = a2002 f11 (f13 (f14 (a))) = (a14·13 )11 = a2002 Hence the answer is (C).

Solutions 2002 AMC 10 P

4

13. (B) If m and w are the current numbers of men and women, respectively, then we have m w m+w + = 1.05 1.20 1.10 or 1 1 1 m 1 · ? = ? . w 1.05 1.10 1.10 1.20 It follows that
m w

=

7 4

so that

m+w w

=

11 4 ,

and

w m+w

=

4 11 .

14. (A) In fact, regradless of the size of angle EID, the area of quadrilateral EIDJ 1 will always be 4 . Let K be the foot of the perpendicular from E to CD and let L be the foot of the perpendicular from E to AD. Then, because right triangle EKI is congruent to right triangle ELJ , the area of quadrilateral EIDJ equals the area of square EKDL which equals 1 4. 15. (D) Consider the set of eight disjoint pairs, {(1,9), (2,10), (3,11), (4,12), (5,13), (6,14), (7,15), (8,16)}. Because only four members of {1, 2, 3, . . . , 20} are not members of these pairs, any subset of size 13 must contain both members of at least one of these pairs and thus must contain two numbers that di?er by 8. The answer is (D) because the twelve element subset {1, 2, 3, 4, 5, 6, 7, 8, 17, 18, 19, 20} contains no two numbers di?ering by 8. 16. (D) Let the ?y be x meters from the ceiling. Then the ?y and point P determine a major diagonal of the rectangular parallelepiped having dimensions 1, 8, and x. Therefore, 12 + 82 + x2 = 92 , and it follows that x = 4.
·2001 17. (C) There are 2002 = 2002 = 1001 · 2001 possible pairs that can be drawn. 2 2·1 10012 2 There are 1001 pairs of di?erent colored marbles, so Pd = 1001 ·2001 . There1000 1 fore, Ps = 1 ? Pd = 2001 , and |Ps ? Pd | = 2001 .

18. (C) Because n3 ? 8n2 + 20n ? 13 = (n ? 1)(n2 ? 7n + 13), for the value to be prime one factor must equal 1 and the other factor must be prime. For n ? 1 = 1 we must have n = 2, and in this case the other factor is the prime 3. So n = 2 is a solution. For n2 ? 7n + 13 = 1, we have n2 ? 7n + 12 = 0 = (n ? 4)(n ? 3), so we must have n = 3 or 4, and in each case the other factor is prime (2 and 3, respectively). Therefore n3 ? 8n2 + 20n ? 13 is a prime for three positive integer values of n. 19. (A) Adding the three equations we obtain (a2 + 6a) + (b2 + 2b) + (c2 + 4c) = ?14, which is equivalent to (a + 3)2 + (b + 1)2 + (c + 2)2 = 0. Therefore a = ?3, b = ?1, c = ?2, and a2 + b2 + c2 = 14.

Solutions 2002 AMC 10 P

5

20. (A) Let U be the set of all three-digit numbers, let S be the set of three-digit numbers that contain no 2s, and let T be the set of three digit numbers that contain no 3s. Then S ∩ T is the set of three-digit numbers containing neither a 2 nor a 3 and U ? (S ∪ T ) is the set of three-digit numbers containing at least one 2 and at least one 3. We have |U | = 900, |S | = |T | = 8 · 92 , and |S ∩ T | = 7 · 82 . Therefore |S ∪ T | = |S | + |T | ? |S ∩ T | = 848, and |U ? (S ∪ T )| = 52. 21. (B) Setting x = 2, and then x = 1001, we have f (2)+2f (1001) = 6 and f (1001)+ 2f (2) = 3003. Subtracting the ?rst equation from twice the second equation we obtain 3f (2) = 6000, so f (2) = 2000. Note that f (x) = 4004 x ? x is such a function. 22. (B) The number of zeros at the end of n! is the largest power of 10 that is a factor of n!. It is also the largest power of 5 that divides n!. In general, the largest power of the prime p in the prime factorization of n! is n n n + 2 + 3 + ··· . 5 5 5 Therefore the largest power of 5 that is a factor of 2002 2002 2002 2002 + + + 5 52 53 54 1000 1000 1000 1000 ?2 + + + 5 52 53 54
2002! (1001!)2

is

= (400 + 80 + 16 + 3) ? 2(200 + 40 + 8 + 1) = 1.

23. (B) We have a?b= Since 12 22 ? 12 32 ? 22 42 ? 32 10012 ? 10002 10012 + + + + ... + ? . 1 3 5 7 2001 2003 =
2k+1 2k+1

(k+1)2 ?k2 2k+1

= 1 for all k ≥ 0, it follows that 10012 2003

a ? b =1 + 1 + ··· + 1? 1001 times

10012 2003 ≈ 1001 ? 500.25 ≈ 500.75. = 1001 ? So the closest integer is 501. 24. (D) Since 12 + 22 + 32 + . . . + 182 > 2002, it follows that n ≤ 17. Then note that 12 + 22 + 32 + . . . + 192 ? 182 ? 122 = 2002, hence n = 17.

Solutions 2002 AMC 10 P

6

25. (D) If there are c (c ≥ 0) correct answers and u (u ≥ 0) unanswered questions and c + u ≤ 25, then the score is 6c + 2.5u. If c is su?ciently large and u is su?ciently small, the same score will be obtained with c ? 5 correct answers and u + 12 unanswered questions (this requires c + u ≤ 18), and also with c ? 10 correct answers and u + 24 unanswered questions. Note that in the latter case we must have c ≥ 10 and c + u ≤ 11. Therefore, for there to be three ways to obtain the score 6c + 2.5u we can only have c = 10 and u = 0, or c = 10 and u = 1, or c = 11 and u = 0. The three such scores are 60, 62.5, and 66, and their sum is 188.5

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