How many ordered pairs of positive integers $(x,y)$ satisfy LCM$(x,y)=400$?

Question

How many ordered pairs of positive integers $(x,y)$ satisfy LCM$(x,y)=400$?

Answer 1

Let's factor $400$ into $2^4\cdot5^2$. This means $x=2^a5^b$ and $y=2^c5^d$ where $max(a,c)=4$ and $max(b,d)=2$. Let’s split into cases:

Case $1$. $a=4$ If $a=4$, then $c$ can be $0,1,2,3,4$.

Subcase $1$. $a=4, b=2$ If $b=2$, $d$ can be $0, 1, 2$, for a total of $5\cdot3=15$ pairs $(x,y)$ in this subcase.

Subcase $2$. $a=4, d=2$ If $d=2$, then $b$ can be $0, 1$ ($2$ was counted in subcase $1$) for a total of $5\cdot2=10$ pairs $(x,y)$ in this subcase.

Case $2$. $c=4$ If $c=4$, then $a$ can be $0, 1, 2, 3$ ($4$ was already counted). Therefore, in this case, subcase $1$ ($c=4, b=2$) and subcase $2$ ($c=4,d=2$) still hold as in case $1$. Thus here in subcase $1$ there are $4\cdot3=12$ pairs and in subcase $2$ there are $4\cdot2=8$ pairs.

Therefore, the total is $15+10+12+8=\boxed{45}$.

Answer 2

$400 = 5^2 2^4 $

$x = 2^ x 5^y , y = 2^a 5^b$

$Max(x,a) = 4, Max(y,b)= 2$

Possibilities of $Max(x,a) = 4$ is $(0,4),(1,4),(2,4),(3,4)(4,4),(4,0),(4,1)(4,2)(4,3)$ = 9

Possibilities of $Max(y,b) = 2$ is $(0,2),(1,2),(2,2),(2,0),(2,1)$ = 5

$\implies 9*5 = 45$