How many pairs?

$(a-200)(b-200) = 40000 = 2^{6}\times 5^{4}$

$40000$ can distribute its factors to the first bracket, and then there's only 1 way to the second bracket. So we can count the number of factors.

Let $a-200 = \pm 2^{k}\times 5^{m}$ , then $b-200 = \pm 2^{6-k}\times 5^{4-m}$

The number of ways = $2\times (7\times 5) = 70$ ways. ( $a,b$ can switch numbers together so that we can multiply by 2)

But for $k=3,m=2, \text{negative sign}$ we get $a = 0, b = 0$ which is impossible.

Therefore, the number of solutions $= 70 - 1 = \boxed{69}$ ~!~!~