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Question 41 — JEE Main 21 Jan 2026 S Ii

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  • 22 September, 2026 at 12:59 am #11670
    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

    22 September, 2026 at 12:59 am #11671
    shahkb4
    Keymaster

    Solution

    \[ \begin{aligned} & \text { Exponent of } 7=\left[\frac{101}{7}\right]+\left[\frac{101}{49}\right]+\left[\frac{101}{49 \times 7}\right]+\cdots \\ & \quad=14+2+0=16 \\ & 101!=7^{16}(\lambda) \text { Answer }(1) \end{aligned} \]
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