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Randomly Chosen Divisors

Compute the probability that a randomly chosen positive divisor of 10^{99} is an integer multiple of 10^{88}

Recall previously that the number of divisors is expressed as \prod_{i}^{k} (e_i + 1) where each e_i is the exponent of the respective prime i in the unique prime factorization of that number. In the previous article, this was established via a simple counting argument on the number of ways of filling out the exponents on the prime factors.

In our problem, it is extremely easy to factor 10^n = 2^n5^n. So 10^{99} = 2^{99}5^{99}, 10^{88} = 2^{88}5^{88}. For a divisor of 10^{99} to be a multiple of 10^{88}, it must be the case that its e_2,e_5 \ge 88. There are 99-88+1 ways of doing this for each exponent, so there are a total of 12 \times 12 divisors of 10^{99} that are multiples of 10^{88}.

Now, there are a total of (99+1)\times(99+1) = 10000 divisors of 10^{99}, hence the probability that a randomly chosen divisor of 10^{99} is also a multiple of 10^{88} is \frac{144}{10000} = \frac{9}{625}. \blacksquare

Posted By Lee