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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

Check the Answer


Answer: is 634.

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.

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