JEE Advanced

Question 42 — JEE Main 24 Jan 2026 S Ii

Viewing 2 posts - 1 through 2 (of 2 total)
  • Author
    Posts
  • #11672
    shahkb4
    Keymaster

    The largest value of n, for which $40^{\mathrm{n}}$ divides 60!, is
    [JEE MAIN 24 Jan 2026 S II]
    (1) 13
    (2) 11
    (3) 12
    (4) 14

    #11673
    shahkb4
    Keymaster

    Solution

    \[ 40=8 \times 5=2^{3} \times 5 \]

    Exponent of 2 in $60!=\left[\frac{60}{2}\right]+\left[\frac{60}{4}\right]+\left[\frac{60}{8}\right]+\left[\frac{60}{16}\right]+\left[\frac{60}{32}\right]+\left[\frac{60}{64}\right]+\cdots$

    \[ =30+15+7+3+1+0+\cdots=56 \]

    Exponent of 5 in $60!=\left[\frac{60}{5}\right]+\left[\frac{60}{25}\right]+\left[\frac{60}{125}\right]+\cdots=12+2+0=14$

    \[ 60!=2^{56} 5^{14}(\mathrm{~K})=2^{14}\left(2^{42}\right) 5^{14} \mathrm{~K}=2^{14}\left(2^{3} \times 5\right)^{14} \mathrm{~K}=2^{14}(40)^{14} \mathrm{~K} \]

    ∴ Maximum value of n is 14
    Answer (4)

Viewing 2 posts - 1 through 2 (of 2 total)
  • You must be logged in to reply to this topic.