It may seem really easy to solve but almost all of us are stuck on this one.

Seems really weird:

Prove that, for all $3\leqslant n$, $S(9^n)\not=9$.

$S(n)$ is the sum of digits of $n$ base $10$.

َAlso, there's no need to check for odd $n$ because in that case the solution is as follows:

  • First of all it's clear that $S(n)$ is divisible by $9$ so if it's not $9$ we're done.
  • Furthermore, if $n$ is odd, the rightmost digit of $9^n$ is $9$, and there's at least one other non-zero digit, so sum of digits of $n$ is more than $9$, and that's it.

But for even $n$ I'm seriously stuck.

  • $\begingroup$ could you clarify what is $S$? $\endgroup$
    – yotam maoz
    Jan 7, 2021 at 16:43
  • $\begingroup$ Yep I'm sorry I'll point it out $\endgroup$
    – Aryan
    Jan 7, 2021 at 16:44
  • 2
    $\begingroup$ it is easy if $n$ is odd (just look at the last digit) $\endgroup$
    – Exodd
    Jan 7, 2021 at 16:52
  • 3
    $\begingroup$ See jstor.org/stable/2695428?seq=1#metadata_info_tab_contents $\endgroup$
    – AnilCh
    Jan 7, 2021 at 18:50
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    $\begingroup$ Try using induction to prove that for any positive integer m , 9^(m+2) cannot have digits summing up to 9 $\endgroup$
    – itp dusra
    Jan 7, 2021 at 18:58

1 Answer 1


Adapting from @ACheca suggested paper "The Decimal Expansions of Powers of 9" by Sapir, Lossers & Montgomery (independently) 1999:

First as stated note that for $\underline{\text{odd }n>1}$, since $9^n\equiv 9 \bmod 10$ we know that the last digit is $9$ and since $9^n> 9$ there are other non-zero digits, so $S(9^n)>9$

Thus for $\underline{\text{even }n}$ we are looking to show that $S(81^k)>9,$ where $\underline{k=n/2>1}$

Let $M(a)$ be the value $(a \bmod 99999)$ with $0\leq M(a)\leq 99999$. Then we can show that $S(M(a))\leq S(a)$ by the following process:

  • Find $M(a)$ from $a$ by taking block of 5 digits and summing the blocks. This process is invariant $\bmod 99999$ and repeated as necessary will produce $M(a)$. However this summing process never increases the digit sum - any carrying effects only reduce the digit sum. Thus $S(M(a))\leq S(a)$.

Now the values of $M(81^k)$ cycle through the following $30$ values: $C =(81,6561,31446,$ $47151,19269,60804,$ $25173,39033,61704,$ $98073,43992,63387,$ $34398,86265,87534,$ $90324,16317,21690,$ $56907,9513,70560,$ $15417,48789,51948,$ $7830,34236,73143,$ $24642,96021,77778)$. By inspection the digit sum of all of these except the first are greater than $9$.


  • For $k \not\equiv 1 \bmod 30$, we have $S(81^k)\geq S(M(81^k))>9$

  • For $k\equiv 1 \bmod 30$ , we have $81^k\equiv 81 \bmod 100$ and recalling that $k>1$ we have $81^k > 81$ and there are other non-zero digits, so for this case also $S(81^k)>9$

... proving the case $n$ even also.

  • 1
    $\begingroup$ @timon92 no, this is specifically looking to generate the mod 99999 values, because those relate to the digit sum. $\endgroup$
    – Joffan
    Jan 7, 2021 at 23:05
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    $\begingroup$ Actually, that's really great. BTW I'll appreciate any solutions without checking 30 values by hand.(If any exists) $\endgroup$
    – Aryan
    Jan 8, 2021 at 3:02
  • $\begingroup$ @AryanHemmati I should go and check my (Excel generated) values against the paper... yes, they match, $\endgroup$
    – Joffan
    Jan 8, 2021 at 3:39

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