Binomial Theorem – Divisibility & Integral Part – JEE Main 28 Jan 2026 Shift 2

Solution:

Statement I:

We rearrange the terms to form pairs:
$$25^{13}+3^{13}+20^{13}+8^{13} = (25^{13}+3^{13}) + (20^{13}+8^{13})$$

We use the property that $x^n + a^n$ is divisible by $(x+a)$ if $n$ is odd.

Here $n=13$ is odd.

$(25^{13}+3^{13})$ is divisible by $25+3=28$.

$(20^{13}+8^{13})$ is divisible by $20+8=28$.

Since both parts are divisible by 28, the entire sum is divisible by 28.

As 28 is a multiple of 7, the expression is divisible by 7.

$\Rightarrow$ Statement I is True.

Statement II:

Let $I$ be the integral part and $f$ be the fractional part of the given expression.
$$(7+4\sqrt{3})^{25} = I + f, \quad \text{where } 0 \le f < 1$$
Let us define a conjugate term $f’$:
$$f’ = (7-4\sqrt{3})^{25}$$

Since $\sqrt{49} > \sqrt{48} \Rightarrow 7 > 4\sqrt{3}$, we have $0 < 7-4\sqrt{3} < 1$.
Thus, $0 < f' < 1$.
Now, add the two expressions:
$$I + f + f’ = (7+4\sqrt{3})^{25} + (7-4\sqrt{3})^{25}$$

Using the binomial expansion $(x+y)^n + (x-y)^n = 2[C_0 x^n + C_2 x^{n-2}y^2 + \dots]$:
$$I + f + f’ = 2 \left[ {}^{25}C_0 7^{25} + {}^{25}C_2 7^{23}(4\sqrt{3})^2 + \dots \right]$$

The term inside the bracket consists of integers, so the RHS is $2 \times (\text{Integer})$, which is an even integer.
$$I + f + f’ = \text{Even Integer}$$

Since $I$ is an integer, $(f+f’)$ must also be an integer.

Checking the range of sum of fractions:
$$0 < f < 1 \quad \text{and} \quad 0 < f' < 1 \Rightarrow 0 < f+f' < 2$$
The only integer between 0 and 2 is 1.
$$f + f’ = 1$$

Substitute this back into the sum equation:
$$I + 1 = \text{Even Integer}$$
$$I = \text{Even Integer} – 1$$

Therefore, $I$ is an odd number.

$\Rightarrow$ Statement II is True.

Both Statement I and Statement II are correct.

Ans. (2)