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Thread: Very large numbers!

  1. #1
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    Very large numbers!

    (3^10)! / 3^x = k

    x and k are integers.

    What is the maximum value for x ?

    I think: (3^10 - 1) / 2 = 29524.

    Sloane's sequence #A003462

    Agree?
    Last edited by DenisB; Oct 26th 2016 at 10:07 AM.
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  2. #2
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    Re: Very large numbers!

    $3^{10}!$ contains all factors $3^k$ up to $k=10$

    the product of these is $3^{55}$

    so $x$ is at least $55$
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  3. #3
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    Re: Very large numbers!

    If we evaluate how many of the the prime factors of 3^10 are 3's, we can start by noting that every third term of the factorial is divisible by three. So there are 3^9 terms that are a divisible by 3.

    Similarly every ninth term is a factor of 9, so there are 3^8 terms that are divisible by 9.

    Keep going. and the total number of 3's that make up the prime factors of 3^10 is 3^9 + 3^8 + 3^ 7 + ...+ 3^1 + 3^0. That seems to equal your answer of (3^10-1)/2, but I'm not sure why....

    Edit - OK, got it. It's not hard to show that:

     \sum _{i=0} ^n k^i = \frac {k^{n+1}-1}{k-1}

    Hence for k=3 and n = 9:

     \sum _{i=0} ^{9} 3^i = \frac {3^{10}-1}{2}
    Last edited by ChipB; Oct 26th 2016 at 01:36 PM.
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  4. #4
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    Re: Very large numbers!

    Denis,
    Your result follows immediately from Legendre's Theorem (Legendre's Theorem - The Prime Factorization of Factorials). This gives the exact power of p, a prime, that divides n factorial.
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