Arjun Suresh (talk | contribs) |
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==={{Template:Author|Happy Mittal|{{mittalweb}} }}=== | ==={{Template:Author|Happy Mittal|{{mittalweb}} }}=== | ||
| − | + | Divisors 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 at least 96 i.e. 96, 97, 98, or 99. | of $10^{99}$. Now Any of those divisors would be a multiple of $10^{96}$ if both a and b are at least 96 i.e. 96, 97, 98, or 99. | ||
So each of a and b have 4 choices each, and so there are 16 divisors which are multiple of $10^{96}$. | So each of a and b have 4 choices each, and so there are 16 divisors which are multiple of $10^{96}$. | ||
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
Divisors 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 at least 96 i.e. 96, 97, 98, or 99.
So each of a and b have 4 choices each, and so there are 16 divisors which are multiple of $10^{96}$.
Thus, required probability
= 16/10000 = 1/625
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
Divisors 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 at least 96 i.e. 96, 97, 98, or 99.
So each of a and b have 4 choices each, and so there are 16 divisors which are multiple of $10^{96}$.
Thus, required probability
= 16/10000 = 1/625
GATECSE