Question

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

• A. $\frac{1}{5}$
• B. $\frac{1}{7}$
• C. $\frac{1}{4}$
• D. $\frac{1}{49}$

Answer 1

Answer: (C)

$\frac{1}{4}$

$7^k$ when divided by $5$, the remainder is:
$2$ if $k=4p+1$ ...(a)
$-1$ if $k=4p+2$ ...(b)
$-2$ if $k=4p+3$ ...(c)
$1$ if $k=4p$ ...(d)
where $p$ is a non-negative integer
Thus, if the remainder in the question has to be $0$:
(i) $m$ and $n$ must be from (a) and (c) or vice versa OR
(ii) $m$ and $n$ must be from (b) and (d) or vice versa.
From $1-100$, there are $25$ numbers of each type of (a), (b), (c) and (d).
Hence, the required probability $=$
$\frac{{(}^{25}{C}_1)\times {(}^{25}{C}_1)\times 2+{(}^{25}{C}_1)\times {(}^{25}{C}_1)\times 2}{{(}^{100}{C}_1)\times {(}^{100}{C}_1)}=\frac{1}{4}$