Q35 / 75 · JEE Main 2026 (28 January Shift 2) · Mathematics
Given below are two statements :
Statement I: $\quad 25^{13}+20^{13}+8^{13}+3^{13}$ is divisible by 7.
Statement II: The integral part of $(7+4 \sqrt{3})^{25}$ is an odd number.
In the light of the above statements, choose the correct answer from the options given below :
Statement I: $\quad 25^{13}+20^{13}+8^{13}+3^{13}$ is divisible by 7.
Statement II: The integral part of $(7+4 \sqrt{3})^{25}$ is an odd number.
In the light of the above statements, choose the correct answer from the options given below :
- AStatement I is true but Statement II is false
- Correct answerBBoth Statement I and Statement II are true
- CStatement I is false but Statement II is true
- DBoth Statement I and Statement II are false
Solution
Statement I:
$25^{13} + 3^{13}$ → divisible by $(25 + 3)$
$20^{13} + 8^{13}$ → divisible by $(20 + 8)$
$\therefore$ divisible by $7$. TRUE
Statement II:
For $(7+4\sqrt{3})^{25}$, let $\alpha = 7+4\sqrt{3}$ and $\beta = 7-4\sqrt{3}$ (conjugate).
Then $\alpha\beta = 1$ and $\alpha + \beta = 14$.
Since $0 < \beta < 1$, we have $\alpha^{25} + \beta^{25}$ is an integer.
The fractional part of $\alpha^{25}$ is $1-\beta^{25}$, making the integral part odd. TRUE.
Both Statement I and Statement II are true.
$25^{13} + 3^{13}$ → divisible by $(25 + 3)$
$20^{13} + 8^{13}$ → divisible by $(20 + 8)$
$\therefore$ divisible by $7$. TRUE
Statement II:
For $(7+4\sqrt{3})^{25}$, let $\alpha = 7+4\sqrt{3}$ and $\beta = 7-4\sqrt{3}$ (conjugate).
Then $\alpha\beta = 1$ and $\alpha + \beta = 14$.
Since $0 < \beta < 1$, we have $\alpha^{25} + \beta^{25}$ is an integer.
The fractional part of $\alpha^{25}$ is $1-\beta^{25}$, making the integral part odd. TRUE.
Both Statement I and Statement II are true.
Hence, the correct answer is option: 'B'.