1988 AIME Problem 5

Attempt Problem 5 of the 1988 AIME below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1988 AIME solutions, or check the answer key.

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5.

Let mn,\frac{m}{n}, in lowest terms, be the probability that a randomly chosen positive divisor of 109910^{99} is an integer multiple of 1088.10^{88}. Find m+n.m+n.

Answer: 634
Concepts:factor countingbasic probabilityprime factorization
Difficulty rating: 1770
Small Hint:

Write every divisor as 2a5b2^a5^b

Big Hint:

Count exponent pairs with a88a\geq88 and b88b\geq88 and reduce the resulting probability

Solution:

Every divisor is 2a5b2^a5^b with 0a990\leq a\leq99 and 0b99,0\leq b\leq99, so there are 1002=10000100^2=10000 divisors. A multiple of 108810^{88} requires 88a9988\leq a\leq99 and 88b99,88\leq b\leq99, giving 122=14412^2=144 divisors. The probability is 14410000=9625,\frac{144}{10000}=\frac{9}{625}, and m+n=9+625=634.m+n=9+625=634.

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