Question

If $m/n$, in lowest terms, be the probability that a randomly chosen positive divisor of $10^{99}$ is an integral multiple of $10^{88}$ then $(m+n)$ is equal to

• A. 634
• B. 643
• C. 632
• D. 692

Answer 1

Answer: (A)

$N=10^{99}=2^{99}\cdot 5^{99}$
$\therefore$ number of divisors of $N=(100)(100)=10^4$
now $10^{88}=2^{88}\cdot 5^{88}$
Hence divisors which are integral multiple of $2^{88}\cdot 5^{88}$ must be of the form of $2^a\cdot 5^b$ where $88\le a,b\le 99$. Thus there are $12\times 12$ ways to choose $a$ and $b$ and hence there are $12\times 12$ divisors which are integral multiple if $2^{88}\cdot 5^{88}$.
Hence $p=\frac{144}{10000}=\frac{9}{625}$
$\therefore m+n=634$