Notice that the prime factorization of [tex]20^{20}[/tex] and [tex]10^{10}[/tex] are [tex]2^{40}\cdot5^{20}[/tex] and [tex]2^{10}\cdot5^{10}[/tex], 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:
[tex]2^y\cdot5^x\cdot2^{10}\cdot5^{10}\\=2^{10+y}\cdot5^{10+x}[/tex]
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:
[tex]\boxed{\frac{341}{861}}[/tex]
also are you, by any chance, the same guest who posted the same question in web2.0?
