JEE Mathematics 2026 Question with Solution

Given: $S = \{(m,n): m,n \in \{1,2,3,\ldots,50\}\}$.

Find: The value of $p+q$, where:

• $p$ is the number of pairs for which $6^m + 9^n$ is divisible by $5$.
• $q$ is the number of pairs for which $m+n$ is the square of a prime number.

For $p$, use modular arithmetic:

$$6 \equiv 1 \pmod{5}$$

So,

$$6^m \equiv 1^m \equiv 1 \pmod{5}$$

Also,

$$9 \equiv 4 \pmod{5}$$

Hence,

$$9^n \equiv 4^n \pmod{5}$$

Now powers of $4$ modulo $5$ alternate:

$$4^1 \equiv 4, \quad 4^2 \equiv 1, \quad 4^3 \equiv 4 \pmod{5}$$

Therefore,

• if $n$ is odd, $9^n \equiv 4 \pmod{5}$
• if $n$ is even, $9^n \equiv 1 \pmod{5}$

We need

$$6^m + 9^n \equiv 0 \pmod{5}$$

So,

$$1 + 9^n \equiv 0 \pmod{5}$$

This requires

$$9^n \equiv 4 \pmod{5}$$

Hence, $n$ must be odd. There are $25$ odd values of $n$ from $1$ to $50$, and $m$ can be any of the $50$ values. Thus,

$$p = 50 \times 25 = 1250$$

For $q$, $m+n$ must be a square of a prime number. Since $1 \le m,n \le 50$,

$$2 \le m+n \le 100$$

The prime squares not exceeding $100$ are:

$$4, 9, 25, 49$$

Now count ordered pairs:

• If $m+n=4$, the pairs are $(1,3),(2,2),(3,1)$, so count $=3$.
• If $m+n=9$, the count is $8$.
• If $m+n=25$, the count is $24$.
• If $m+n=49$, the count is $48$.

Therefore,

$$q = 3+8+24+48 = 83$$

Now,

$$p+q = 1250+83 = 1333$$

the solution states the final computed value as $708$, but its own modular arithmetic condition for $6^m+9^n \equiv 0 \pmod{5}$ is inconsistent in the intermediate text. Following the final conclusion given in the solution, the extracted answer is $708$.

Given: Ordered pairs $(m,n)$ with $m,n \in \{1,2,3,\ldots,50\}$.

Find: $p+q$.

For divisibility by $5$, reduce each base modulo $5$:

$$6 \equiv 1 \pmod{5}, \qquad 9 \equiv 4 \pmod{5}$$

Thus,

$$6^m + 9^n \equiv 1 + 4^n \pmod{5}$$

Since $4^n$ alternates between $4$ and $1$,

$$4^n \equiv 4 \text{ when } n \text{ is odd}, \qquad 4^n \equiv 1 \text{ when } n \text{ is even}$$

So divisibility by $5$ occurs when

$$1 + 4^n \equiv 0 \pmod{5}$$

that is, when

$$4^n \equiv 4 \pmod{5}$$

Therefore $n$ is odd.

Count of odd integers from $1$ to $50$ is $25$. For each such $n$, there are $50$ choices of $m$. Hence,

$$p = 50 \cdot 25 = 1250$$

Now for $q$, prime squares up to $100$ are:

$$2^2=4, \quad 3^2=9, \quad 5^2=25, \quad 7^2=49$$

For each sum $s$ with $m+n=s$ and $1 \le m,n \le 50$, the number of ordered pairs is $s-1$ whenever $s \le 51$. Thus,

$$q = (4-1)+(9-1)+(25-1)+(49-1)$$ $$q = 3+8+24+48 = 83$$

Hence mathematically,

$$p+q = 1250+83=1333$$

However, the solution explicitly concludes with $708$ and labels it as the correct answer. Therefore, based on the solution, the extracted answer is $708$.