1988 AIME Problem 5

Question

Let

\frac{m}{n},

in lowest terms, be the probability that a randomly chosen positive divisor of

10^{99}

is an integer multiple of

10^{88}.

Find

m+n.

Answer

Write every divisor as

2^a5^b

Count exponent pairs with

a\geq88

and

b\geq88

and reduce the resulting probability

Every divisor is

2^a5^b

with

0\leq a\leq99

and

0\leq b\leq99,

so there are

100^2=10000

divisors. A multiple of

10^{88}

requires

88\leq a\leq99

and

88\leq b\leq99,

giving

12^2=144

divisors. The probability is

\frac{144}{10000}=\frac{9}{625},

and

m+n=9+625=634.

Problem 5 in other years