Question

If m/n ,in lowest terms, be the probability that a randomly chose positive divisor of ${10}^{99}$ is an integral multiple of ${10}^{88}$ than find the value of $(m+n)$.

Answer 1

${10}^{99}=\left(2{)}^{99}\right(5{)}^{99}$

To we multiple of ${10}^{88}$

we need to take $\{2^{88}.....{.2}^{99}\}$ and $\{5^{88}......{.5}^{99}\}$

there are $12\times 12=144$

Probability $=\frac{144}{(100{)}^2}$

$=\frac{144}{10000}=\frac{9}{625}$

$m=9n=626$

$m+n=634$