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

Step 1: Understanding the Question:
We are given a set \(S\) consisting of all ordered pairs \((m,n)\), where \[ 1 \le m \le 50 \quad \text{and} \quad 1 \le n \le 50. \] Thus, the total number of ordered pairs is \(50 \times 50 = 2500\).
We define:

• \(p\): the number of pairs for which \(6^m + 9^n\) is a multiple of 5,
• \(q\): the number of pairs for which \(m+n\) is the square of a prime number.

We are required to compute \(p+q\).
Step 2: Key Approach:

• For \(p\), we use modular arithmetic to analyze \(6^m + 9^n \pmod{5}\).
• For \(q\), we identify all prime squares between 2 and 100 and count the valid ordered pairs \((m,n)\) whose sum equals each such value.

Step 3: Detailed Explanation:
Part 1: Calculation of \(p\)
We want: \[ 6^m + 9^n \equiv 0 \pmod{5}. \] Working modulo 5: \[ 6 \equiv 1 \pmod{5} \quad \Rightarrow \quad 6^m \equiv 1 \pmod{5} \text{ for all } m. \] \[ 9 \equiv 4 \equiv -1 \pmod{5} \quad \Rightarrow \quad 9^n \equiv (-1)^n \pmod{5}. \] Thus, the condition becomes: \[ 1 + (-1)^n \equiv 0 \pmod{5}. \] If \(n\) is even: \[ 1 + (-1)^n = 2 \not\equiv 0 \pmod{5}. \] If \(n\) is odd: \[ 1 + (-1)^n = 0 \equiv 0 \pmod{5}. \] Hence, the condition is satisfied if and only if \(n\) is odd.
There are 50 choices for \(m\) and 25 odd values of \(n\) in \(\{1,2,\dots,50\}\).
Therefore, \[ p = 50 \times 25 = 1250. \]
Part 2: Calculation of \(q\)
We require: \[ m + n = k^2, \] where \(k\) is a prime number and \(1 \le m,n \le 50\). The possible sums range from: \[ 2 \le m+n \le 100. \] Prime numbers whose squares lie in this interval are: \[ 2^2 = 4, \quad 3^2 = 9, \quad 5^2 = 25, \quad 7^2 = 49. \] For each sum \(S\), the number of valid ordered pairs \((m,n)\) is \(S-1\), since all values lie within the allowed range.

• \(m+n = 4 \Rightarrow 3\) pairs
• \(m+n = 9 \Rightarrow 8\) pairs
• \(m+n = 25 \Rightarrow 24\) pairs
• \(m+n = 49 \Rightarrow 48\) pairs

Hence, \[ q = 3 + 8 + 24 + 48 = 83. \]
Part 3: Final Computation
\[ p + q = 1250 + 83 = 1333. \]
Step 4: Final Answer:
The value of \(p + q\) is: \[ \boxed{1333}. \]