Question

Two numbers $m$ and $n$ are randomly selected from among the divisors of $1000.$ Assuming each divisor is equally likely to be chosen, the probability that $m$ and $n$ are co-prime and $\text{LCM}(m,n)$ is $1000$ is

• A. $\frac{3}{16}$
• B. $\frac{1}{64}$
• C. $\frac{1}{4}$
• D. $\frac{1}{16}$

Answer 1

The correct option is B $\frac{1}{64}$

$1000=2^3\times 5^3$
Number of factors of $1000=(3+1)\times (3+1)=16$
So, total number of pairs $(m,n)$ is $16\times 16=256$

$m,n$ are co-prime.
$\Rightarrow \text{HCF}(m,n)=1$
We know that
$\text{HCF}(m,n)\times \text{LCM}(m,n)=m\times n$
So, $\text{LCM}(m,n)=m\times n$

The possible pairs are $(1,1000),(1000,1),(125,8),(8,125)$
$\therefore$ Required probability $=\frac{4}{256}=\frac{1}{64}$