Let S=(m, n): m, n 1,2,3, .

Question

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 ,...

Answer: 1333

Accepted Answer

Since $6 \equiv 1 \pmod 5$, we have $6^m \equiv 1 \pmod 5$ for all $m$. Since $9 \equiv -1 \pmod 5$, we have $9^n \equiv (-1)^n \pmod 5$.

So $6^m + 9^n \equiv 1 + (-1)^n \pmod 5$, which is $0 \pmod 5$ exactly when $n$ is odd (giving $1-1=0$).

Thus $m$ can be any of $50$ values, and $n$ must be one of the $25$ odd values in $\{1,\ldots,50\}$:
$$p = 50 \times 25 = 1250$$

For $q$: we need $m+n$ to equal the square of a prime, with $m,n \in \{1,\ldots,50\}$, so $m+n \in \{2,\ldots,100\}$. The squares of primes in this range are $4, 9, 25, 49$ (since $11^2=121>100$).

For a fixed sum $s = m+n$ with $2 \le s \le 51$, the number of ordered pairs $(m,n)$ is $s-1$.
- $s=4 \Rightarrow 3$ pairs
- $s=9 \Rightarrow 8$ pairs
- $s=25 \Rightarrow 24$ pairs
- $s=49 \Rightarrow 48$ pairs

$q = 3+8+24+48 = 83$

$p+q = 1250 + 83 = 1333$

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