JEE Mathematics 2026 Question with Solution

Given: Two statements are to be checked.

• Statement I: $25^{13}+20^{13}+31^{13}$ is divisible by $7$
• Statement II: the integral part of $\left(7+4\sqrt3\right)^{25}$ is odd

Find: Which option correctly describes the truth values of the two statements.

For Statement I, work modulo $7$.

$$25\equiv4,\quad 20\equiv6,\quad 31\equiv3\pmod{7}$$

Using Fermat's theorem,

$$a^6\equiv1\pmod7$$

so

$$a^{13}=a^{12}a=(a^6)^2a\equiv a\pmod7$$

Hence,

$$25^{13}+20^{13}+31^{13}\equiv 4+6+3=13\equiv0\pmod7$$

Therefore, Statement I is true.

For Statement II, use the conjugate.

$$(7+4\sqrt3)(7-4\sqrt3)=49-48=1$$

Hence,

$$(7+4\sqrt3)^{25}+(7-4\sqrt3)^{25}\in\mathbb{Z}$$

Also,

$$0<7-4\sqrt3<1$$

so

$$(7-4\sqrt3)^{25}\in(0,1)$$

Thus, the integral part of $\left(7+4\sqrt3\right)^{25}$ is

$$(7+4\sqrt3)^{25}+(7-4\sqrt3)^{25}-1$$

and this is odd. Therefore, Statement II is true.

So, both Statement I and Statement II are true. The correct option is D.

Given: The expression $\left(7+4\sqrt3\right)^{25}$ appears in Statement II.

Find: A quick way to decide whether its integral part is odd.

Observe that $7+4\sqrt3$ and $7-4\sqrt3$ are conjugates and

$$(7+4\sqrt3)(7-4\sqrt3)=1$$

Therefore,

$$(7+4\sqrt3)^{25}+(7-4\sqrt3)^{25}$$

is an integer. Since $0<7-4\sqrt3<1$, its $25$th power lies between $0$ and $1$. So subtracting that small positive quantity from the integer shows that the floor of $\left(7+4\sqrt3\right)^{25}$ is exactly one less than that integer, hence odd.

This shortcut works because the conjugate term is positive but less than $1$, so it determines the floor immediately without expanding any power.