What is the maximum value of k for which 105!/12k is an integer?​

Question

What is the maximum value of k for which 105!/12k is an integer?​

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Peyton 1 month 2021-10-27T02:52:16+00:00 2 Answers 0 views 0

Answers ( )

    0
    2021-10-27T02:53:21+00:00

    The max value of k is 50

    Step-by-step explanation:

    We know that for n!, and a given prime number p, the highest power of p in n! is given by

    \boxed{[\frac{n}{p}]+[\frac{n}{p^2}]+[\frac{n}{p^3}]+.....}

    Where [ ] is the greatest integer function

    In the given question

    \frac{105!}{12^k}

    =\frac{105!}{(3\times 2^2)^k}

    =\frac{105!}{3^k\times 2^{2k}}

    Power of 3 in 105!

    =[\frac{105}{3}]+[\frac{105}{3^2}]+[\frac{105}{3^3}]+[\frac{105}{3^4}+[\frac{105}{3^5}+...

    =[35]+[11.67]+[3.89]+[1.3]+[0.43]

    =35+11+3+1+0

    =50

    Power of 2 in 105!

    =[\frac{105}{2}]+[\frac{105}{2^2}]+[\frac{105}{2^3}]+[\frac{105}{2^4}+[\frac{105}{2^5}+\frac{105}{2^6}+\frac{105}{2^7}+...

    =[52.5]+[26.25]+[13.125]+[6.5625]+[3.28125]+[1.640625]+[0.8203125]

    =52+26+13+6+3+1+0

    =101

    Thus, if we take k=50, \frac{105!}{12^k} will be an integer

    Hope this answer is helpful.

    Know More:

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    0
    2021-10-27T02:53:39+00:00

    Answer:

    50

    Step-by-step explanation:

    Highest power of 3 in 105! =

    [105/3^1]+[105/3^2]+[105/3^3]+[105/3^4]

    = 35+11+3+1 => 50

    where [ ] this bracket signifies greatest integer function.

    As heighest power of 2 will be greater in 105! We don’t need to calculate this for all the prime factors of 12 to get the maximum value of k for 105!/12^k to be an integer

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