Aspire Faculty ID #18111 · Topic: JEE Main 2026 (28 January Evening Shift) · Just now
JEE Main 2026 (28 January Evening Shift)

Given below two statements :

Statement I : $ 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 :

Solution

Statement I :

$ 25^{13} + 3^{13} \quad\text{and}\quad 20^{13} + 8^{13} $

divisible by $ (25 + 3) $ and divisible by $ (20 + 8) $

$ \therefore $ divisible by $ 7 $

Statement II :

Let $ R = (7 + 4\sqrt{3})^{25} = I + f $

$ R' = (7 - 4\sqrt{3})^{25} = f' $

$ \therefore R + R' = I + f + f' = 2^{25} C_0 7^{25} + 25C_2 7^{23}(4\sqrt{3})^2 + \ldots $

$ I + f + f' = \text{even integer} $

$ \Rightarrow I = \text{odd integer} $

$ \therefore 0 < f + f' < 2 \Rightarrow f + f' = 1 $

$ \therefore $ Both the statements are correct

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