Arjun Suresh (talk | contribs) |
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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 | 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 | + | 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 | + | So each of a and b have 4 choices each, and so there are 16 divisors which are multiple of $10^{96}$. |
<br> | <br> | ||
Thus, required probability | Thus, required probability | ||
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[[Category: GATE2010]] | [[Category: GATE2010]] | ||
[[Category: Probability questions]] | [[Category: Probability questions]] | ||
| + | [[Category:Mathematics Questions from GATE]] | ||
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
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 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
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 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