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?

yes, gavin is right.

step-by-step explanation:

gavin is correct. 

this problem is about converting 9% per year to how many percent per quarter. the rate is still the same. it only differed in the time period. so, through dimensional analysis,

9% per year  × (1 year / 4 quarters) = 2.25% per quarter

if he was asked to convert 9% per year to how many percent per quarter  compounded quarterly, then you will not simply divide by four. this is because compounding is not linear.

hope this , have a good day. c;