Question

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

• A. $\frac{1}{4}$
• B. $\frac{1}{7}$
• C. $\frac{1}{8}$
• D. $\frac{1}{49}$

Answer 1

Answer: (C)

$7^m+7^n=\left[(5+2{)}^m+(5+2{)}^n\right]\equiv 5\times$ integer $+2^m+2^n$
$\therefore 7^m+7^n$ is divisible by 5 iff $2^m+2^n$ is divisible by $5$ and
so unit place of $2^m+2^n$ must be 0 since it cannot be $5$ . $m$ possible $n$
$13,7,11,15,....=25$
$24,8,12,.....=25$
$31,5,9,.....=25$
$42,6,10,....=25$
Since $2^1+2^3\equiv 2^3+2^1$ so $(1,3)$ and $(3,1)$ are same as favourable cases.
$\therefore$required probability$=\frac{25\times 50}{100\times 100}=\frac{1}{8}$