INTRODUCING 5 - days-a-week problem solving session for Math Olympiad and ISI Entrance. Learn More 

May 17, 2020

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




Check the Answer

Answer: is 634.

AIME I, 1988, Question 5

Elementary Number Theory by David Burton

Try with Hints

First hint


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 =\(\frac{m}{n}=\frac{144}{10000}=\frac{9}{625}\)


Subscribe to Cheenta at Youtube

Leave a Reply

This site uses Akismet to reduce spam. Learn how your comment data is processed.

Cheenta. Passion for Mathematics

Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.