# Positive divisor | AIME I, 1988 | Question 5

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1988 based on Positive divisor.

## Positive Divisors- AIME I, 1988

Let $\frac{m}{n}$ in lowest term, be the probability that a randomly chosen positive divisor of $10^{99}$ is an integer multiple of $10^{88}$. Find m+n.

• is 107
• is 634
• is 840
• cannot be determined from the given information

### Key Concepts

Integers

DIvisors

Algebra

AIME I, 1988, Question 5

Elementary Number Theory by David Burton

## Try with Hints

First hint

$10^{99}=2^{99}5^{99}$

or, (99+1)(99+1)=10000 factors

Second Hint

those factors divisible by $10^{88}$

are bijection to number of factors $10{11}=2^{11}5^{11}$ has, which is (11+1)(11+1)=144

Final Step

one required probability =$\frac{m}{n}=\frac{144}{10000}=\frac{9}{625}$

m+n=634.

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