Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1988 based on Positive divisor.
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.
Integers
DIvisors
Algebra
Answer: is 634.
AIME I, 1988, Question 5
Elementary Number Theory by David Burton
\(10^{99}=2^{99}5^{99}\)
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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