Question

Let S=\{(m, n): m, n \(\in\) \{1, 2, 3, ....., 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 of 50x50 = 2500 ordered pairs (m, n).
We need to find the number of pairs `p` that satisfy the condition that \(6^m+9^n\) is a multiple of 5.
We also need to find the number of pairs `q` that satisfy the condition that `m+n` is the square of a prime number.
Finally, we need to calculate the sum `p+q`.
Step 2: Key Formula or Approach:
For p: We will use modular arithmetic. The condition is \(6^m+9^n \equiv 0 \pmod{5}\). We will analyze the powers of 6 and 9 modulo 5 to simplify the condition.
For q: We will first identify all the numbers between 2 (1+1) and 100 (50+50) that are squares of prime numbers. Then, for each such sum S, we will count the number of integer pairs (m, n) such that \(m+n=S\) and \(1 \le m, n \le 50\).
Step 3: Detailed Explanation:
Part 1: Calculation of p
We need to find when \(6^m+9^n\) is a multiple of 5. Let's work in modulo 5.
\[ 6 \equiv 1 \pmod{5} \implies 6^m \equiv 1^m \equiv 1 \pmod{5} \text{ for any } m \ge 1.
\] \[ 9 \equiv 4 \equiv -1 \pmod{5} \implies 9^n \equiv (-1)^n \pmod{5}.
\] The condition becomes:
\[ 1 + (-1)^n \equiv 0 \pmod{5}
\] This means \(1 + (-1)^n\) must be a multiple of 5.
If n is even, \(1 + (-1)^n = 1 + 1 = 2\), which is not a multiple of 5.
If n is odd, \(1 + (-1)^n = 1 - 1 = 0\), which is a multiple of 5.
So, the condition is satisfied if and only if `n` is an odd number.
The value of `m` can be any integer from 1 to 50, so there are 50 choices for `m`.
The value of `n` must be an odd integer from 1 to 50. These are {1, 3, 5, ..., 49}. The number of such values is 25.
The total number of pairs (m, n) is \(p = (\text{choices for m}) \times (\text{choices for n}) = 50 \times 25 = 1250\).
Part 2: Calculation of q
We need \(m+n = k^2\), where k is a prime number and \(1 \le m, n \le 50\).
The sum \(S=m+n\) can range from \(1+1=2\) to \(50+50=100\).
The prime numbers whose squares fall in this range are 2, 3, 5, and 7.
- \(k=2 \implies k^2 = 4\).
- \(k=3 \implies k^2 = 9\).
- \(k=5 \implies k^2 = 25\).
- \(k=7 \implies k^2 = 49\).
The next prime is 11, with \(11^2 = 121>100\).
Now, we count the number of pairs (m, n) for each sum S:
- For \(m+n = 4\): The pairs are (1,3), (2,2), (3,1). All values are in the range [1, 50]. This gives 3 pairs.
- For \(m+n = 9\): The pairs are (1,8), (2,7), ..., (8,1). All values are in range. This gives 8 pairs.
- For \(m+n = 25\): The pairs are (1,24), (2,23), ..., (24,1). All values are in range. This gives 24 pairs.
- For \(m+n = 49\): The pairs are (1,48), (2,47), ..., (48,1). All values are in range. This gives 48 pairs.
The total count for q is the sum of these counts:
\[ q = 3 + 8 + 24 + 48 = 83.
\] Part 3: Calculation of p + q
\[ p+q = 1250 + 83 = 1333.
\] Step 4: Final Answer:
The value of p + q is 1333.