Question 31 — JEE Main 21 Jan 2026 (S I)

Question

Let $\mathrm{S}=\{(\mathrm{m}, \mathrm{n}): \mathrm{m}, \mathrm{n} \in\{1,2,3, \ldots \ldots, 50\}\}$. 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 ( $\mathrm{m}, \mathrm{n}$ )
in S such that $\mathrm{m}+\mathrm{n}$ is a square of a prime number is q , then $\mathrm{p}+\mathrm{q}$ is equal to $\_\_\_\_$
[JEE MAIN 21 Jan 2026 S I]

Solution

$6^{\mathrm{m}}+9^{\mathrm{n}}=(5+1)^{\mathrm{m}}+(10-1)^{\mathrm{n}}$
$=\left({ }^{m} \mathrm{C}_{0} 5^{m}+\ldots+{ }^{m} \mathrm{C}_{m-1} 5+{ }^{m} \mathrm{C}_{m}\right)+\left({ }^{n} \mathrm{C}_{0} 10^{n}-{ }^{n} \mathrm{C}_{1} 10^{n-1}+\cdots+{ }^{n} \mathrm{C}_{n-1} 10+{ }^{n} \mathrm{C}_{n}(-1)^{n}\right]$
$=(5 \mathrm{k}+1)+\left(10 \lambda+(-1)^{\mathrm{n}}\right)$
$=5(\mathrm{k}+2 \lambda)+1+(-1)^{\mathrm{n}}$
$m \in\{1,2,3,4, \ldots, 49,50\}$
n must be an odd no i.e. $\mathrm{n} \in\{1,3,5,7, \ldots, 49\}$;
p $=50 \times 25=1250$
$\mathrm{m}+\mathrm{n}=2^{2}=4$
$(\mathrm{m}, \mathrm{n})=(1,3),(2,2),(3,1) \rightarrow 3$
$m+n=3^{2}=9$
$(\mathrm{m}, \mathrm{n})=(1,8),(2,7)(3,6), \ldots(8,1) \rightarrow 8$
$\mathrm{m}+\mathrm{n}=5^{2}=25$
$(\mathrm{m}, \mathrm{n})=(1,24)(2,23),(3,21) \ldots(24,1) \rightarrow 24$
$\mathrm{m}+\mathrm{n}=7^{2}=49$
$(m, m)=(1,48)(2,47),(3,46), \ldots(48,1) \rightarrow 48$
$(\mathrm{m}+\mathrm{n})_{\text {max }}=100$
$\mathrm{m}+\mathrm{n} \neq 11^{2}=121$
$\mathrm{q}=3+8+24+48=83$
$\therefore p+q=1250+83=1333$ Answer(1333)