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. $\frac{5}{18}$
• B. $\frac{1}{7}$
• C. $\frac{1}{8}$
• D. $\frac{1}{49}$

Answer 1

The correct option is A $\frac{5}{18}$

We know $7^k,k\in N$, has $1,3,9,7$ at the units place for $k=4p,4p-1,4p-2,4p-3$ respectively,
where $p=1,2,3,...$

Clearly $7^m+7^n$ will be divisible by $5$ if $7^m$ has $3$ or $7$ in the units place and

$7^n$ has $7$ or $3$ in units place or $7^m$ has $1$ or $9$ in the units place and $7^n$ has $1$ or $9$ in the units place.

$\therefore$ For any choice of $m,n$ the digits in the units place of $7^m+7^n$ is $2,4,6,0$0r$8$.

It is divisible by $5$ only when this digit is $0$.

$\therefore$ Required probability $=\frac{1}{5}$