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 \(q\), then \(p + q\) is equal to

Hint:

For problems involving divisibility by a small number, modular arithmetic is an extremely powerful and fast tool.
Simplify the bases modulo the divisor before considering the exponents.
For counting problems like finding `q`, be systematic. List all possible cases and count for each case carefully, ensuring the variables stay within their defined ranges.

Answer 1

Correct Answer: 6

To solve the problem, we begin by addressing the set \(S = \{(m,n) : m,n \in \{1,2,3,\ldots,50\}\}\), which contains \(50 \times 50 = 2500\) total elements.

1. **Finding \(p\):** We need \(6^m + 9^n \equiv 0 \pmod{5}\). Calculate \(6^m \mod 5\) and \(9^n \equiv (4^n) \mod 5\):

• \(6 \equiv 1 \pmod{5}\), so \(6^m \equiv 1^m \equiv 1 \pmod{5}\).
• \(9 \equiv 4 \pmod{5}\), observing powers of 4 mod 5: \(4^1 \equiv 4\), \(4^2 \equiv 1\), \(4^3 \equiv 4\) and so forth.

Thus, \(9^n \equiv 1 \pmod{5}\) if \(n\) is even and \(9^n \equiv 4 \pmod{5}\) if \(n\) is odd. Thus, \(6^m + 9^n \equiv 0 \pmod{5}\) when both \(m\) and \(n\) are even. For \(m, n \in \{2, 4, 6, \ldots, 50\}\), there are 25 choices for each, hence \(p=25 \times 25 = 625\).

2. **Finding \(q\):** Determine pairs \((m,n)\) such that \(m+n\) is a square of a prime number. The squares of prime numbers ≤ 100 are \(4, 9, 25, 49\). For each:\(m+n=4\): Possible pairs: \((1,3), (2,2), (3,1)\) — 3 pairs.
\(m+n=9\): Pairs: \((1,8), (2,7), (3,6), (4,5), (5,4), (6,3), (7,2), (8,1)\) — 8 pairs.
\(m+n=25\): Pairs from \((1,24)\) to \((24,1)\) — 24 pairs.
\(m+n=49\): Pairs from \((1,48)\) to \((48,1)\) — 48 pairs.
Hence, \(q=3+8+24+48=83\).

3. **Calculate and confirm \(p+q\):** \(p + q = 625 + 83 = 708\). The expected range was \(6,6\), but computing reveals \(708\) is the solution which fits the precise calculation, not the range inadvertently mentioned.