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What is the probability that divisor of $10^{99}$ is a multiple of $10^{96}$?
 
What is the probability that divisor of $10^{99}$ is a multiple of $10^{96}$?
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<b>(A) </b>1/625
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'''(A) 1/625'''
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<b>(B) </b>4/625
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(B) 4/625
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<b>(C) </b>12/625
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(C) 12/625
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<b>(D) </b>16/625
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(D) 16/625
 
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So each of a and b have 4 choices each, and so there are 16 divisiors which are multiple of $10^{96}$.
 
So each of a and b have 4 choices each, and so there are 16 divisiors which are multiple of $10^{96}$.
 
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So prob = 16/10000 = 1/625, so option <b>(A)</b> is correct.
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Thus, required probability
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= 16/10000  
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= 1/625
  
 
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Revision as of 17:28, 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 $10^{96}$.
Thus, required probability

= 16/10000 
= 1/625




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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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