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Question 41 — JEE Main 21 Jan 2026 S Ii

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  • 3 June, 2026 at 8:23 pm #1109
    shahkb4
    Keymaster

    The largest $\mathrm{n} \in \mathrm{N}$, for which $7^{\mathrm{n}}$ divides 101!, is
    [JEE MAIN 21 Jan 2026 S II]
    (1) 16
    (2) 18
    (3) 15
    (4) 19

    3 June, 2026 at 8:23 pm #1110
    shahkb4
    Keymaster

    Solution

    Exponent of $7=\left[\frac{101}{7}\right]+\left[\frac{101}{49}\right]+\left[\frac{101}{49 \times 7}\right]+\cdots$

    \[ =14+2+0=16 \]

    $101!=7^{16}(\lambda)$ Answer(1)

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