So Many Constraints!

Let $a=kb+r$ , Where $k, r \in \mathbb{Z}\$$ and $0 \leq r < b$ . Since $\gcd(a, b)=1$ , We have $\gcd(r, b)=1$ .

Now the third equation becomes $\frac{r}{b}k=2013$ , So $b \mid rk$ . Since $\gcd(r,b)=1$ , We get $b \mid k$ . Let $k=lb, l \in \mathbb{Z}$ , so $rl=2013$ .

Now $r \mid 2013$ and $r < b \leq 100$ , So $r=1, 3, 11, 33, 61$ .

At this point, note that each choice of $r, b$ such that $r < b \leq 100, r \mid 2013$ and $\gcd(r, b)=1$ corresponds to exactly one solution for $(a, b)$ . (Since $a=kb+r=lb^2+r=\frac{2013}{r}b^2+r$ )

If $r=1$ , We have $1 < b \leq 100, \gcd(b, 1)=1$ , Giving $99$ solutions for $b$ .

If $r=3$ , We have $3 < b \leq 100, \gcd(b, 3)=1$ , Giving $97-32=65$ solutions for $b$ .

If $r=11$ , We have $11 < b \leq 100, \gcd(b, 11)=1$ , Giving $89-8=81$ solutions for $b$ .

If $r=33$ , We have $33 < b \leq 100, \gcd(b, 33)=1$ , Giving $67-22-6+2=41$ solutions for $b$ .

If $r=61$ , We have $61 < b \leq 100, \gcd(b, 61)=1$ , Giving $39$ solutions for $b$ .

This gives a Total of $325$ solutions for $(r,b)$ and hence for $(a, b)$ .