If the probability that a randomly chosen positive divisor of ${10}^{99}$ is an integer multiple of ${10}^{88}$ is $\frac{m}{n}$, (m,n are relatively prime natural numbers) then the number of digits in $m+n=$

${10}^{99}=2^{99}{5}^{99}$, So it has $(99+1)(99+1)=10000$ factors. Out of these, we only want those factors of ${10}^{99}$ which are divisible by ${10}^{88}$; it is easy to draw a bijection to the number of factors that ${10}^{11}=2^{11}{5}^{11}$ has, which is $(11+1)(11+1)=144.$ Our probability is $\frac{m}{n}=\frac{144}{10000}=\frac{9}{625},$ and  $m+n=634.$