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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)

best_answer

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 .

 

Posted by

HARSH KANKARIA

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