Q46 / 75 · JEE Main 2026 (21 January Shift 1) · Mathematics
Let $\mathrm{S}=\{(\mathrm{m}, \mathrm{n}): \mathrm{m}, \mathrm{n} \in\{1,2,3, \ldots.., 50\}\}$. If the number of elements $(\mathrm{m}, \mathrm{n})$ in S such that $6^{\mathrm{m}}+9^{\mathrm{n}}$ 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 $\mathrm{p}+\mathrm{q}$ is equal to $\_\_\_\_$.
Correct answer: 1333
Solution
For $6^m + 9^n \equiv 0 \pmod{5}$:
$6 \equiv 1 \pmod{5}$ and $9 \equiv -1 \pmod{5}$
So $6^m + 9^n \equiv 1 + (-1)^n \pmod{5}$
This equals 0 only when n is odd.
Number of odd n in {1,...,50} = 25, and m can be any of 50 values.
$p = 50 \times 25 = 1250$
For $m + n$ = square of prime, possible values are 4, 9, 25, 49.
$m + n = 4$: 3 pairs
$m + n = 9$: 8 pairs
$m + n = 25$: 24 pairs
$m + n = 49$: 48 pairs
$q = 3 + 8 + 24 + 48 = 83$
$p + q = 1250 + 83 = 1333$
$6 \equiv 1 \pmod{5}$ and $9 \equiv -1 \pmod{5}$
So $6^m + 9^n \equiv 1 + (-1)^n \pmod{5}$
This equals 0 only when n is odd.
Number of odd n in {1,...,50} = 25, and m can be any of 50 values.
$p = 50 \times 25 = 1250$
For $m + n$ = square of prime, possible values are 4, 9, 25, 49.
$m + n = 4$: 3 pairs
$m + n = 9$: 8 pairs
$m + n = 25$: 24 pairs
$m + n = 49$: 48 pairs
$q = 3 + 8 + 24 + 48 = 83$
$p + q = 1250 + 83 = 1333$
Hence, the correct answer is 1333.