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  1. #1
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    number theroy

    Determine all positive integers n so that n^2 equals the sum of the factorial of the digit of n.
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  2. #2
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    by "the sum of factorials of n" i mean n! + (n - 1)! + (n - 2)! + ... + 2! + 1
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    Well, factorials grow at a larger rate than exponentials for large values of n. In fact, for n>=4, n, the sum of the factorials will always be larger than their exponents, so you can just do the other cases by hand. To prove that it is always larger, you can do induction, starting with n =4
    4^2 =- 16
    \sum_{1 4} n! = 24 + 6 + 2 +1 = 33
    and show for 5, assume true for n, and show true for n +1.

    Then you can look at 1, 2, 3.
    1! = 1^2 = 1
    2! = 2+1=3 != 2^2 = 4
    3! = 6 + 2 + 1=9 = 3^2 = 9

    So only 1 and 3.
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    Quote Originally Posted by terr13 View Post
    Well, factorials grow at a larger rate than exponentials for large values of n. In fact, for n>=4, n, the sum of the factorials will always be larger than their exponents, so you can just do the other cases by hand. To prove that it is always larger, you can do induction, starting with n =4
    4^2 =- 16
    \sum_{1 4} n! = 24 + 6 + 2 +1 = 33
    and show for 5, assume true for n, and show true for n +1.

    Then you can look at 1, 2, 3.
    1! = 1^2 = 1
    2! = 2+1=3 != 2^2 = 4
    3! = 6 + 2 + 1=9 = 3^2 = 9

    So only 1 and 3.
    what do u mean by \sum_{1 4} n! = 24 + 6 + 2 +1 = 33?? Cant tell what you did here
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  5. #5
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    I'm taking the sum of that expression from 1 to 4, so 1! + 2! + 3! + 4! = 1 + 2 + 6 + 24 = 33, which is larger than 4^2 = 16. This is always true for n>=4.
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    Senior Member vincisonfire's Avatar
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    The sum of the factorial of digit of 1000 is 4 but 10^3^2=10^6. It is not about the sum of factorial from 1 to n.
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    Senior Member vincisonfire's Avatar
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    I know that 71 works.
    71^2 = 1! + 7! = 5041
    But I don't know how to list these solutions
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    Quote Originally Posted by vincisonfire View Post
    I know that 71 works.
    71^2 = 1! + 7! = 5041
    But I don't know how to list these solutions
    71 does not work
    it would have to be 71^2=1! + 2! +3! + 4!+....71! 71! it self is just bigger then 71^2
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  9. #9
    Senior Member vincisonfire's Avatar
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    The question says that n^2 should be equal to the sum of the factorial of the DIGIT of n.
    n = a1*10^k+...+an.
    You search n such that (a1*10^k+...+an)^2 = a1! + ...+ an!
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  10. #10
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    Quote Originally Posted by vincisonfire View Post
    The question says that n^2 should be equal to the sum of the factorial of the DIGIT of n.
    n = a1*10^k+...+an.
    You search n such that (a1*10^k+...+an)^2 = a1! + ...+ an!
    This is the idea:
    n = sum a_k * 10^k, 0 <= k <= m
    with a_k in {0, 1, 2, .., 9} and a_m > 0. When is

    n^2 = sum (a_k)!.

    For example,
    7! + 1! = 71^2.

    I'm asking to find (with proof) all such examples.

    Does this help?
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