SMO 2010 Open Section, Round 1

$100(a+b)=ab-100$

$ab-100(a+b)+10000=10100$

$(a-100)(b-100)=10100$

Observe that 10100 is either the product of 2 negative factors or 2 positive factors.

$a-100\geq -99$ and $b-100\geq -99$ for $a,b$ to be both positive integers.

The maximum value of $(a-100)(b-100)$ is 9801 for both $a-100$ and $b-100$ to be negative factors, but as $(a-100)(b-100)=10100$ , there are no solutions for both factors to be negative.

So the only solutions are for both factors to be positive.

$10100 = 2^2 \cdot 5^2 \cdot 101^1$

So there are $3\cdot 3 \cdot 2 = 18$ positive factors of 10100. Hence there are 18 pairs of positive integers $(a,b)$