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Revision as of 17:23, 15 April 2014

What is the probability that divisor of $10^{99}$ is a multiple of $10^{96}$?
(A) 1/625   (B) 4/625   (C) 12/625   (D) 16/625

Solution by Happy Mittal

Divisiors of $10^{99}$ are of the form $2^a*5^b$, where a and b can go from 0 to 99 each, so there are 10000 divisors of $10^{99}$. Now Any of those divisors would be a multiple of $10^{96}$ if both a and b are atleast 96 i.e. 96, 97, 98, or 99. So each of a and b have 4 choices each, and so there are 16 divisiors which are multiple of $10^{96}$.
So prob = 16/10000 = 1/625, so option (A) is correct.




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What is the probability that divisor of $10^{99}$ is a multiple of $10^{96}$?
(A) 1/625   (B) 4/625   (C) 12/625   (D) 16/625

Solution by Happy Mittal[edit]

Divisiors of $10^{99}$ are of the form $2^a*5^b$, where a and b can go from 0 to 99 each, so there are 10000 divisors of $10^{99}$. Now Any of those divisors would be a multiple of $10^{96}$ if both a and b are atleast 96 i.e. 96, 97, 98, or 99. So each of a and b have 4 choices each, and so there are 16 divisiors which are multiple of $10^{96}$.
So prob = 16/10000 = 1/625, so option (A) is correct.




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