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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

## Check the Answer

But try the problem first…

Answer: is 634.

Source

Suggested Reading

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.

## Other useful links

- https://www.cheenta.com/rational-number-and-integer-prmo-2019-question-9/
- https://www.youtube.com/watch?v=lBPFR9xequA

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