I doubt anyone can get this but... 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 $10^{10}$?

Notice that the prime factorization of and are and , respectively. also, notice that both of their prime factorizations contain only 2 and 5.

Let the divisor of 20^20 that is a multiple of 10^10 be:

where y and x are positive integers.

We can have y equal to 0, 1, 2, ... 30 before the exponent of 2 exceeds 40, and we can have x equal to 0, 1, 2, ... 10 before the exponent of 5 exceeds 20.

That is 11*31= 341 numbers in total.

There are (40+1)(20+1)=861 factors in 20^20, which means that the final answer is:

also are you, by any chance, the same guest who posted the same question in web2.0?

go hook jvc hgv cj xk jvc 6756

30

step-by-step explanation:

times by times