Question

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

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

Answer 1

The correct option is A $\frac{1}{4}$

$7^1=7,7^2=49,7^3=343,7^4=2401,\cdots$
Therefore, for $7^r,rϵN$ the number ends with 7, 9, 3, 1, 7, . . .
$\therefore 7^m+7^n$ will be divisible by 5 if it ends in 0 or 5.
The last digit of m and n belongs to the set {1,3,7,9}
Hence there are $4\times 4=16$ combinations of digits which determines the last digit of the sum.
The favourable cases among them are { (1,9),(9,1),(3,7),(7,3) }

Hence, the required probability is
$P=\frac{4}{16}=\frac{1}{4}$