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

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

Since, $m$ and $n$ are selected between $1$ and $100$, hence sample space $=100\times 100$
Also, $7^1=1,7^2=49,7^3=343,7^4=2401$
$7^5=16807$ etc.
Hence, $1,3,7$ and $9$ will be the last digits in the power of $7$. Hence, for favourable cases.
$\underset{\downarrow }{nm}$
$1,21,21,3....1,100$
$2,12,22,3...2,100$
$100,1100,2100,3...100,100$
for $m=1,n=3,7,11,....97$
Therefore, favorable cases $=25$
for $m=2,n=4,8,12,10,.....100$
Therefore, favorable cases $=25$
Similarly for every $m$, favourable $n$ are $25$.
Total favourable cases $=100\times 25$
Hence, required probability $=\frac{100\times 25}{100\times 100}=\frac{1}{4}$.