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