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.
• B.
• C.
• D.

Answer 1

The correct option is A

We know that $7^1=7,7^2=49,7^3=343,7^4=2401,7^5=16807$. Therefore, $7^k$ (where $k\in Z$) results in a number whose unit’s digit is 7 or 9 or 3 or 1.
Now, $7^m+7^n$ will be divisible by 5 if unit’s place digit in the resulting number is 5 or 0. Clearly, it can never be 5. But it can be 0 if we consider values of m and n such that the sum of unit’s place digits become 0. And this can be done choosing
$\begin{array}{c}m=1,5,9,.....,97\\ n=3,7,11,....,99\end{array}\}$(25 options each)[7+3=10]
Or
$\begin{array}{c}m=2,6,10,.....,98\\ n=4,8,12,....,100\end{array}\}\left(25optionseach\right)[9+3=10]$
Therefore, the total number of selections of m, n such that $7^m+7^n$ is divisible by 5 is
$(25\times 25+25\times 25)\times 2$ (since we can interchange values of m and n).
Also the number of total possible selections of m and n out of 100 is $100\times 100$. Therefore, the required probability is
$\frac{2(25\times 25+25\times 25)}{100\times 100}=\frac{1}{4}$