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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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$\Sigma_d (n!) = O(n \ln (n))$ grows faster than $n$, so there's no clear reason to think it's asymptotically true. Have you looked for counterexamples? –  G. H. Faust Jul 27 at 9:57
    
@G.H.Faust - I've tried up to $n=100$ on my pc, but that clearly wasn't enough :). –  R B Jul 27 at 10:14
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what about sum of binary digits? (following the usual observation that there's nothing special about 10) –  Mitch Jul 27 at 17:41
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@Mitch, I wrote a program to generate the first counterexamples in arbitrary bases. For bases 2 through 30, the first counterexamples are 10, 43, 86, 87, 188, 156, 291, 364, 432, 410, 7, 510, 4, 4, 4, 813, 4, 1079, 4, 1900, 6, 10, 6, 2330, 2147, 5, 3463, 2401 and 7 respectively. –  G. H. Faust Jul 28 at 10:04
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@G.H.Faust: Add it to OEIS!! –  Mitch Jul 28 at 12:06

1 Answer 1

up vote 41 down vote accepted

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 at 13:19
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One of the classics "patterns that eventually fail". –  Hashir Omer Jul 27 at 16:45
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Curious that $432, 532, 632$ are each counterexamples (though not $732$) –  Henry Jul 27 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 at 8:00

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