Let S = \(m, n): m, n in \1, 2, 3, dots, 50\ \$S = \{(m, n): m, n \in \{1, 2, 3, \dots, 50\} \}$ . If the number of elements (m, n)$(m, n)$ in S$S$ such that 6^m + 9^n$6^m + 9^n$ is a multiple of 5 is p$p$ and the number of elements (m, n)$(m, n)$ in S$S$ such that m + n$m + n$ is a square of a prime number is q$q$ , then p + q$p + q$ is equal to......

### Related Formula Modular arithmetic reductions for power cycles: a equiv b pmodm Rightarrow a^k equiv b^k pmodm$$a \equiv b \pmod{m} \Rightarrow a^k \equiv b^k \pmod{m}$$ ### Core Logic Analyze condition p$p$: (6^m + 9^n)$(6^m + 9^n)$ is divisible by 5. 6 equiv 1 pmod 5 Rightarrow 6^m equiv 1^m equiv 1 pmod 5$6 \equiv 1 \pmod 5 \Rightarrow 6^m \equiv 1^m \equiv 1 \pmod 5$. 9 equiv -1 pmod 5 Rightarrow 9^n equiv (-1)^n pmod 5$9 \equiv -1 \pmod 5 \Rightarrow 9^n \equiv (-1)^n \pmod 5$. For the sum to be divisible by 5: 1 + (-1)^n equiv 0 pmod 5 Rightarrow (-1)^n = -1$1 + (-1)^n \equiv 0 \pmod 5 \Rightarrow (-1)^n = -1$. This implies n$n$ must be an ODD integer. Since m in \1, 2, dots, 50\$m \in \{1, 2, \dots, 50\}$, m$m$ can be anything (50 choices). Since n$n$ must be odd in \1, dots, 50\$\{1, \dots, 50\}$, n$n$ has 25 choices. p = 50 times 25 = 1250$p = 50 \times 25 = 1250$. ### Step 1: Compute q Analyze condition q$q$: (m + n)$(m + n)$ is the square of a prime number. Max value of m+n = 50+50 = 100$m+n = 50+50 = 100$. Primes whose squares are leq 100$\leq 100$: 2, 3, 5, 7$2, 3, 5, 7$. Their squares are 4, 9, 25, 49$4, 9, 25, 49$. So m+n$m+n$ can be 4, 9, 25, 49$4, 9, 25, 49$. Match List-I with List-II: Explanation for counts: If m+n = S$m+n = S$, and m, n geq 1$m, n \geq 1$, the number of ways is S-1$S-1$ (since S leq 50$S \leq 50$). For S=4$S=4$: 3 ways. For S=9$S=9$: 8 ways. For S=25$S=25$: 24 ways. For S=49$S=49$: 48 ways. q = 3 + 8 + 24 + 48 = 83$q = 3 + 8 + 24 + 48 = 83$. ### Step 2: Final Sum p + q = 1250 + 83 = 1333$$p + q = 1250 + 83 = 1333$$ ### Pattern Recognition Modular exponentiation immediately shrinks large powers to pm 1$\pm 1$. The sum m+n=S$m+n=S$ where 1 le m,n le N$1 \le m,n \le N$ has exactly S-1$S-1$ solutions if S le N$S \le N$, allowing instant combinatorics tallying without manual counting. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Permutations and Combinations