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Denote by $\Sigma_d(t)$ the sum of digits in the decimal representation of the number $t$.

Prove / disprove:

$$\forall n\in \mathbb N:\ \ \Sigma_d (n!) | n!$$

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1 Answers1

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It's not true. The first counterexample is for $ n = 432 $. The sum of the digits in $ 432! $ is 3897, which you can see using Wolfram Alpha. But the prime factorisation of 3897 is $ 3^2 \times 433 $, so $ 432! $ cannot be divisible by its sum of digits.

The list of counterexamples is sequence A066419 in the OEIS.

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    The fact that the first counterexample has $n + 1$ as its only large prime factor is somewhat remarkable. It's like the hole-in-one of counterexamples. – Ryan Reich Jul 27 '14 at 13:19
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    One of the classics "patterns that eventually fail". – Hashir Omer Jul 27 '14 at 16:45
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    Curious that $432, 532, 632$ are each counterexamples (though not $732$) – Henry Jul 27 '14 at 20:34
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    @HashirOmer: I wouldn't say that. n! is divisible by 9 for n >= 6, therefore the sum of digits is divisible by 9. You would expect that the sum of digits is eventually 9 times a prime, and since the sum of digits grows faster than n, you would expect that eventually that prime is greater than n. Very predictable that it would happen. – gnasher729 Jul 28 '14 at 08:00