<s>[INST] What are the number of ordered pairs $(x,y)$ where both $x$ and $y$ divide $20^{19}$, however $xy$ doesn't? What are the number of ordered pairs $(x,y)$ where both $x$ and $y$ divide $20^{19}$, however $xy$ doesn't?
I started by taking the prime factorization of $20^{19}$ to get : $2^{38}5^{19}$.
I then noticed, that the only way for $xy$ to not divide this number was if the powers of $x$ and $y$ add to greater than $38$ or $19$. For example, if $x=2^{38}$ and $y = 2^{2}$, then $xy$ would be $2^{40}$ which would not divide $2^{38}5^{19}$ evenly. 
My question is, what would be the total number of ordered pairs?
 [/INST] Well, If $x = 2^a5^c$ and $y = 2^b5^d$
thenn we have $a+b > 38$ or $c+d > 19$.
To have $a+b > 38$ while $a \le 38$ and $b \le 38$ if we have $a = k$ then $b$ can be as small as $39-k$ and can be as large as $38$.  Those are $k$ options.  $(38+1) - (39-k) = k$.  So there are $\sum_{k=1}^{38}k = \frac {38*39}2 = 741$ such pairs.
And likes there are $\sum_{k=1}^{19} k = 190$ such possible pairs.
If $(a,b)$ is such a pairc $c,d$ can be anything from $0$ to $19$.  Thatn is $20^2 = 400$ options.
So there are $741*400$ pairs where $a+b > 38$.
Like wise there are $190*39^2$ pairs were $c + d > 19$.
But we counted all the pairs where $a+b > 38$ and $c+d >19$ twice!  So we must subtract that number of pairs.  There are $190*741$ so pairs.
So the total number of pairs where $a+b>38$ or $c+d > 19$ or both is
$741*400+190*39^2 - 190*741$.
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