# Prove that the sum of the digits of $9^n$ is not equal to $9$ for $n \ge 3$

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.

• could you clarify what is $S$? Jan 7, 2021 at 16:43
• Yep I'm sorry I'll point it out Jan 7, 2021 at 16:44
• it is easy if $n$ is odd (just look at the last digit) Jan 7, 2021 at 16:52
• Jan 7, 2021 at 18:50
• Try using induction to prove that for any positive integer m , 9^(m+2) cannot have digits summing up to 9 Jan 7, 2021 at 18:58

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 $$9$$.

Thus:

• 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.

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