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If $S_n = 1! + 2! + 3! + \dots + n!$, is there any term in $S_n$ which is a perfect cube or out of $S_1$, $S_2$, $S_3$, $\dots S_n$ is there any term which is a perfect cube, where $n$ is any natural number.

Michael Albanese
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john fedric
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    I have tried solving this problem by calculating last 2 digits in sum of factorial series and found that it is always coming out to be 13 for n>10 , but i cannot infer anything from this.. – john fedric Oct 08 '13 at 12:56
  • @MichaelAlbanese ,Cubes of Numbers 17, 117, 217 ....have 13 as its last two digits.. – john fedric Oct 08 '13 at 13:07
  • That is a good idea. Unfortunately, $17^3$ ends in $13$. It takes three digits to succeed. If this works for you, you could answer your own question. – Ross Millikan Oct 08 '13 at 13:08
  • Yeah, I realised my mistake a little too late. – Michael Albanese Oct 08 '13 at 13:08
  • @RossMillikan, i can also calculate the last three digits for n>15, but how to proceed after that..?? – john fedric Oct 08 '13 at 13:10
  • One more thing if this sum ever a perfect cube then it would be a perfect cube of number having its unit digit 7, because unit digit in that case would be 3.. – john fedric Oct 08 '13 at 13:19
  • It is not as simple as I thought. It stabilizes at 313, but 217^3 ends in 313. So now we know that if $n \gt 15, n=217+1000k$ for some $k$ I think there must be some other base than 10 that will work for this. – Ross Millikan Oct 08 '13 at 13:19
  • Actually it is 313+1000k for n>15 – john fedric Oct 08 '13 at 13:25
  • The sum is 313+1000k, but the cube root (if it existed) would be 217+1000k' – Ross Millikan Oct 08 '13 at 13:31
  • My answer is essentially your idea of checking the last digits, but in a base that works. I tried powers of $2$ a bit, but that didn't work, then the next prime hit the jackpot. – Ross Millikan Oct 08 '13 at 13:36

2 Answers2

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All factorials above $8!$ have a factor of $27$ and $S_8 \equiv 9 \pmod {27}$ As there is no solution to $k^3 \equiv 9 \pmod {27}$, we cannot have $n \ge 9$. Then just checking $n$ up through $8$, only $S_1=1$ is a perfect cube.

Ross Millikan
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Alternatively, we can also use $\pmod{7}$ to prove that $1!+2!+3!+\cdots+n!$ is never a perfect cube $\forall n\geq2$.

Note that $7\mid n!$ if $n\geq7$, so whenever $n\geq6$, then $1!+2!+3!+4!+5!+6!\equiv5\pmod{7}$.

But all perfect cubes are either equal to $0,\pm1\pmod{7}$, so $1!+2!+3!+\cdots n!$ isn’t a perfect cube if $n\geq6$.

We can check the values of $n$ between $2$ to $6$ and by brute force, we see that the sum is not a perfect cube.

Bonus:

Note that $11\mid1!+2!+3!+\cdots+10!$, so the sum(if $n\geq10$) is divisible by $11$, and so, the sum(if $n\geq21$) must be divisible by $121$ in order to be a perfect cube.

In this answer, it is said that $11^{2}\mid n!$ if $n!\geq22$, but because $1!+2!+3!+\cdots+21!\not\equiv0\pmod{11^{2}}$, then the sum is not divisible by $11^{2}$, and as a result, not a perfect cube.

Thirdy Yabata
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