Question

If the integers m and n are chosen at random from 1 to 100, then the probability that a number of the form $7^n+7^m$ is divisible by 5, equal to

• A. $\frac{1}{4}$
• B. $\frac{1}{2}$
• C. $\frac{1}{8}$
• D. $\frac{1}{3}$

Answer 1

The correct option is A $\frac{1}{4}$

Let $I=7^n+7^m$, then we observe that $7^1,7^2,7^3$ and $7^4$ ends in $7,9,3$ and $1$ respectively.

Thus, $7^1$ ends in 7, 9, 3 or 1 according as i is of the form 4k + 1, 4k + 2, 4k - 1 and 4k respectively.

If S is the sample space, then $n\left(S\right)=(100{)}^2$
$7^m+7^n$ is divisible by 5, if
(i) m is of the form 4k + 1 and n is of the form 4k - 1 or
(ii) m is of the form 4k + 2 and n is of the form 4k or
(iii) m is of the form 4k - 1 and n is of the form 4k + 1 or
(iv) m is of the form 4k and n is of the form 4k + 2 or

So, number of favourable ordered pairs $(m,n)=4\times 25\times 25$

$\therefore$ Required probability $=\frac{4\times 25\times 25}{(100{)}^2}=\frac{1}{4}$.

Hence, option A is correct.