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 $5$ , equals to

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

Answer 1

Answer: (A)

$\frac{1}{4}$

Answer 2

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 $4k + 1, 4k+2, 4k-1$, respectively.
If $S$ is the sample space, then $n(S) = (100)^2$
$7^m + 7^n$ 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 $4k + 1$
Thus, number of favourable ordered pairs
$(m, n)=4 \times 25 \times 25$
Hence, required probability
$=\frac{4 \times 25 \times 25}{(100)^{2}}=\frac{1}{4}$