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

Answer: (C)

$\frac{1}{4}$

$7^m$, $m\in N$ ends in $7,9,3or1$

If $m=4k+1$, where $k=\{0,1,2,3,...\}$, then $7^m$ ends in $7$........ category$\left(1\right)$

If $m=4k+2$, where $k=\{0,1,2,3,...\}$, then $7^m$ ends in $9$......... category$\left(2\right)$

If $m=4k+3$, where $k=\{0,1,2,3,...\}$, then $7^m$ ends in $3$......... category$\left(3\right)$

If $m=4k$, where $k=\{1,2,3,...\}$, then $7^m$ ends in $1$......... category$\left(4\right)$

From $1$ to $100$ there are $25$ numbers in each category.

Total number of ways of choosing $m$ and $n$ is equal to $100\times 100$

For a number $7^m+7^n$ to be divisible by $5$, we take $(m,n)$ combination from $($category$\left(1\right)$, category$\left(3\right)$ $)$ or from $($category$\left(2\right)$, category$\left(4\right))$

Hence, favorable number of ways$=2(25\times 25+25\times 25)$ (taking into account order from two categories)

Hence, probability$=\frac{4(25\times 25)}{100\times 100}$

$=\frac{1}{4}$