Question

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

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

Answer 1

Answer: (A)

$7^1=7,7^2=49,7^3=343,7^4=2401,...$
Therefore, for $7^r,r\in N$ the number ends at unit place $7,9,3,1,7$, ...
$\therefore 7^m+7^n$ will be divisible by $5$ if it end at $5$ or $0$
But it cannot end at $5$.
Also for end at $0$.
For this $m$ and $n$ should be as follows

For any given value of $m$, there will be $25$ values of $n$.
Hence, the probability of the required event is
$\frac{100\times 25}{100\times 100}=\frac{1}{4}$
NOTE Power of prime numbers have cyclic numbers in their unit place.