We choose a positive divisor of 20^20 at random (with all divisors equally likely to be chosen). What is the probability that we chose a multiple of 10^18?

Guest Feb 16, 2022

#1**+6 **

20^20 prime factorized is 2^40 x 5^20. Thus, using the prime factorization trick to find the # of factors, 20^20 has 861 factors.

10^18 prime factorszed is 2^18 x 5^18. Thus, using the prime factorization trick to find the # of factors, 10^18 has 361 factors.

Since all factors greater than 10^18 in 20^20 are a multiple of 10^18, thus there are 861 - 361 + 1 = 501 factors that are a multiple of 10^18 when choosing one from 20^20.

(The reason I had a "+ 1" is because one factor of 10^18 is 10^18, which is also a multiple of 10^18. Thus there are 360 factors that do not work in 20^20.)

That means, the probability of choosing a divisor from 20^20 and having it be a multiple of 10^18 is 501/861.

Simplifying, **we get the probability as 167/287**.

proyaop Feb 16, 2022