jmc

The identity $\mathrm{lcm}[a,b]\cdot \gcd (a,b)=ab$ holds for all positive integer pairs $(a,b)$, so in this case, we have $$13200=\mathrm{lcm}[r,100]\cdot \gcd (r,100)=r\cdot 100.$$Solving this equation yields $r=132$, so we are looking for $\mathrm{lcm}[132,100]$. We have the prime factorizations $132=2^2\cdot 3\cdot 11$ and $100=2^2\cdot 5^2$, so, taking the maximum exponent of each prime, we obtain $$\mathrm{lcm}[132,100]=2^2\cdot 3\cdot 5^2\cdot 11=(2^2\cdot 5^2)(3\cdot 11)=(100)(33)=\boxed{3300}.$$(We could also note that the common prime factors of $132$ and $100$ are just $2^2$, which tells us that $\gcd (132,100)=4$ and so $\mathrm{lcm}[132,100]=\frac{13200}{4}=3300$.)