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# Positive divisor | AIME I, 1988 | Question 5

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1990 based on combinatorics in Tournament.

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

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

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

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

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

m+n=634.