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 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.
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 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.
GATECSE