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  • The number of ordered pairs <img alt="(m, n), m, n \in\{1$, $2, \ldots, 50\}" src="https://entrancecorner.oncodecogs.com/gif.latex?%28m%2C%20n%29%2C%20m%2C%20n%20%5Cin%5C%7B1%24%2C%20%242%2C%20%5Cl

The number of ordered pairs (m, n), m, n \in\{1$, $2, \ldots, 50\} such that \mathrm{6^m+9^n} is a multiple of 5 is

Option: 1

1250


Option: 2

2500


Option: 3

500


Option: 4

 625


Answers (1)

best_answer

As the last digit of \mathrm{6^{m}, m \in \mathbf{N}} is \mathrm{6,6^{m}+9^{n}} will be divisible by 5 if the unit's digit of \mathrm{9^{n}}  is 4 or 9. This is possible when \mathrm{n} is odd.

\mathrm{\therefore \quad}  required number of ordered pairs \mathrm{=50 \times 25=1250}\mathrm{=50 \times 25=1250}.

Posted by

Ritika Kankaria

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