To find the required sum, \( p+q \), we need to evaluate two conditions for elements \( (m,n) \) in the set \( S = \{ (m,n) : m,n \in \{1, 2, 3, \dots, 50\} \} \).
Step 1: Calculate \( p \), the number of elements where \( 6^m + 9^n \equiv 0 \pmod{5} \).
Since \( 6 \equiv 1 \pmod{5} \) and \( 9 \equiv 4 \pmod{5} \), we have:
- \(6^m \equiv 1^m \equiv 1 \pmod{5}\).
- \(9^n \equiv 4^n \pmod{5}\) follows the cycle \( (4,1) \), repeating every 2 terms.
We need \( 1 + 4^n \equiv 0 \pmod{5} \rightarrow 4^n \equiv -1 \equiv 4 \pmod{5} \).
This holds when \( n \) is odd since the cycle of \( 4^n \) is: \(4, 1, 4, 1, \ldots\).
Thus, for 25 odd values of \( n \), \( m \) can be any of the 50 values from 1 to 50.
Therefore, \( p = 50 \times 25 = 1250 \).
Step 2: Calculate \( q \), the number of elements where \( m+n \) is a square of a prime number.
- Squares of prime numbers ≤ 100 (max sum of \( m, n \)): \( 2^2=4 \), \( 3^2=9 \), \( 5^2=25 \), \( 7^2=49 \).
Count solutions for each square:
| Square | Solutions |
| 4 | 3 (pairs: (1,3), (2,2), (3,1)) |
| 9 | 8 (pairs: (1,8), (2,7), ..., (8,1)) |
| 25 | 24 (pairs from (1,24) to (24,1)) |
| 49 | 48 (pairs from (1,48) to (48,1)) |
Thus, \( q = 3 + 8 + 24 + 48 = 83 \).
Final Calculation:
\( p + q = 1250 + 83 = 1333 \).
This falls within the given range [1333,1333]. Hence, the final value is \( \boxed{1333} \).
Similar Questions
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- Let $\alpha = \frac{-1 + i\sqrt{3}}{2}$ and $\beta = \frac{-1 - i\sqrt{3}}{2}$, $i = \sqrt{-1}$. If $(7 - 7\alpha + 9\beta)^{20} + (9 + 7\alpha - 7\beta)^{20} + (-7 + 9\alpha + 7\beta)^{20} + (14 + 7\alpha + 7\beta)^{20} = m^{10}$, then $m$ is _________
Let S be the set of the first 11 natural numbers. Then the number of elements in $A = \{B \subseteq S : n(B) \geq 2$ and the product of all elements of B is even is ________.
- The number of $3 \times 2$ matrices A, which can be formed using the elements of the set $\{-2, -1, 0, 1, 2\}$ such that the sum of all the diagonal elements of $A^T A$ is 5, is ______
- The number of numbers greater than 5000, less than 9000 and divisible by 3, that can be formed using the digits 0, 1, 2, 5, 9, if the repetition of the digits is allowed, is ______
- Let $P = [p_{ij}]$ and $Q = [q_{ij}]$ be two square matrices of order 3 such that $q_{ij} = 2^{(i + j - 1)} p_{ij}$ and $\det(Q) = 2^{10}$. Then the value of $\det(\text{adj}(\text{adj } P))$ is:
