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 $5$ is

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

Answer 1

The correct option is A $\frac{1}{5}$

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 unit place and $7^n$ has $7$ or $3$ in the inits place or $7^m$ has $1$ or $9$ in the unit place and $7^n$ has $9$ or $1$ in the unit place.
$\therefore$ For any choice of $m,n$ the digit in the units place of $7^m+7^n$ is $2,4,6,0$ or $8$.
It is divisible by $5$ only when this digit is $0$.
$\therefore$ required probability $=\frac{1}{5}$