The number of ordered pairs <img alt="\mathrm{(m, n), m, n \in\{1,2, \ldots, 100\}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%28m%2C%20n%29%2C%20m%2C%20n%20%5Cin%5C%7B1%2C2

The number of ordered pairs $\mathrm{(m, n), m, n \in\{1,2, \ldots, 100\}}$ such that $\mathrm{7^{m}+7^{n}}$ is divisible by 5 is
Option: 1
1250

Option: 2
2000

Option: 3
2500

Option: 4
5000

Answers (1)

Note that $\mathrm{7^{r}(r \in N)}$ ends in $7,9,3 \: or \: 1$ (corresponding to $r=1,2,3$ and 4 respectively.)

Thus, $\mathrm{7^{m}+7^{n}}$ cannot end in 5 for any values of $\mathrm{m, n \in N}$ .

In other words, for $\mathrm{7^{m}+7^{n}}$ to be divisible by 5 , it should end in 0 .
For $\mathrm{7^{m}+7^{n}}$ o cnd in 0 , the forms of $\mathrm{m}$ and $\mathrm{n}$ should be as follows:

$\begin{tabular}{|l|l|l|} \hline & $m$ & $n$ \\ \hline 1 & $4 r$ & $4 s+2$ \\ \hline 2 & $4 r+1$ & $4 s+3$ \\ \hline 3 & $4 r+2$ & $4 s$ \\ \hline 4 & $4 r+3$ & $4 s+1$ \\ \hline \end{tabular}$

Thus, for a given value of $m$ there are just 25 values of $n$ for which $7^{m}+7^{n}$ ends in 0 . [For instance, if $m=4 r$ , then $n=2,6,10, \ldots, 98$

$\therefore \quad$ there are $100 \times 25=2500$ ordered pairs $(m, n)$ for which $7^{m}+7^{n}$ is divisible by 5 .