The number of ordered pairs (m, n),where m, n | vedclass

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(A) We need $6^m + 9^n \equiv 0 \pmod{5}$.Since $6 \equiv 1 \pmod{5}$,we have $6^m \equiv 1^m \equiv 1 \pmod{5}$ for all $m \in \{1, 2, \ldots, 50\}$.

Vedclass · Accepted Answer

$1250$

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(A) We need $6^m + 9^n \equiv 0 \pmod{5}$.Since $6 \equiv 1 \pmod{5}$,we have $6^m \equiv 1^m \equiv 1 \pmod{5}$ for all $m \in \{1, 2, \ldots, 50\}$.Since $9 \equiv -1 \pmod{5}$,we have $9^n \equiv (-1)^n \pmod{5}$.Thus,$6^m + 9^n \equiv 1 + (-1)^n \pmod{5}$.For the expression to be a multiple of $5$,we require $1 + (-1)^n \equiv 0 \pmod{5}$,which implies $(-1)^n \equiv -1 \pmod{5}$.This holds if and only if $n$ is an odd integer.In the set $\{1, 2, 3, \ldots, 50\}$,there are $50$ possible values for $m$ and $25$ odd values for $n$ (i.e.,$\{1, 3, 5, \ldots, 49\}$).Therefore,the total number of ordered pairs $(m, n)$ is $50 	imes 25 = 1250$.

The number of ordered pairs $(m, n)$,where $m, n \in \{1, 2, 3, \ldots, 50\}$,such that $6^m + 9^n$ is a multiple of $5$ is

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