Let S = {(m, n): m, n in {1, 2, 3, ldots, 50} | vedclass

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(A) For $p$: $6^m + 9^n \equiv 1^m + (-1)^n \equiv 1 + (-1)^n \pmod{5}$.For this to be a multiple of $5$,$1 + (-1)^n \equiv 0 \pmod{5}$,which implies

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$1333$

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(A) For $p$: $6^m + 9^n \equiv 1^m + (-1)^n \equiv 1 + (-1)^n \pmod{5}$.For this to be a multiple of $5$,$1 + (-1)^n \equiv 0 \pmod{5}$,which implies $(-1)^n = -1$.This occurs when $n$ is odd. In the set $\{1, 2, \ldots, 50\}$,there are $25$ odd values for $n$ and $50$ values for $m$.Thus,$p = 50 	imes 25 = 1250$.For $q$: $m + n$ must be a square of a prime number. The possible squares of primes are $2^2 = 4$,$3^2 = 9$,$5^2 = 25$,and $7^2 = 49$.$m + n = 4$$3$ pairs: $(1,3), (2,2), (3,1)$$m + n = 9$$8$ pairs: $(1,8), \ldots, (8,1)$$m + n = 25$$24$ pairs: $(1,24), \ldots, (24,1)$$m + n = 49$$48$ pairs: $(1,48), \ldots, (48,1)$$q = 3 + 8 + 24 + 48 = 83$.$p + q = 1250 + 83 = 1333$.

Let $S = \{(m, n): m, n \in \{1, 2, 3, \ldots, 50\}\}$. If the number of elements $(m, n)$ in $S$ such that $6^{m} + 9^{n}$ is a multiple of $5$ is $p$ and the number of elements $(m, n)$ in $S$ such that $m + n$ is a square of a prime number is $q$,then $p + q$ is equal to :

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