# For every $b$ in the power $a^{b}$, does there exist an $a$ such that the digit sum of this power is equal to $a$?

$$1^0 = 1\to 1 =1$$

$$x^1=x\to x=x\;\forall x$$.

$$9^2 = 81\to 8+1=9$$

$$8^3=512\to 5+1+2=8$$.

$$7^4=2401\to 2+4+0+1=7$$

$$46^5 = 205962976\to 2+0+5+9+6+2+9+7+6=46$$

$$64^6 = 68719476736\to 6+8+7+1+9+4+7+6+7+3+6 = 64$$

$$68^7= 6722988818432\to 6+7+2+2+9+8+8+8+1+8+4+3+2 = 68$$

$$54^8 = 72301961339136\to 7+2+3+0+1+9+6+1+3+3+9+1+3+6=54$$

$$71^9 = 45848500718449031$$
$$\downarrow$$
$$4+5+8+4+8+5+0+0+7+1+8+4+4+9+0+3+1 = 71$$

Conjecture:

Given a positive integer $$b$$, there exists a positive integer $$a$$ such that the digit sum of $$a^b$$ is equal to $$a$$.

Can this be proven? I don't know how to prove it; it was about 3:45am in the morning and I couldn't go to sleep, so I just went on my calculator and messed around because I was bored. That's when I noticed this cool property and decided to share it here.

It is now 4:35am so... I gotta go to bed. I'll see you in some hours and hopefully gather the time to work on this. Sorry about that!

Oh, incidentally, I also noticed that the digit sum of $$29^5$$ is $$23$$ and the digit sum of $$23^5$$ is $$29$$, so... there are cycles here. Same for $$31$$ and $$34$$. Also, the digit sum of $$13^2$$ is $$16$$ and the digit sum of $$16^2$$ is $$13$$. These cycles seem to only have two numbers involved, but I think regarding the seventh power, there is more than two involved (start with $$72^7$$ I think), however there is also a cycle between two numbers of seventh powers (between $$44$$ and $$62$$). Does this help? I don't know.

I have to go to bed. Good night!

• @user477343 This looks like an interesting problem. However, as currently stated, the conjecture is always trivially true for $a = 1$. I assume you mean for $a \gt 1$. – John Omielan Apr 19 at 19:58
• There is the trivial solution $1^b=1$, so perhaps the question needs to be stated more precisely. – Steven Clark Apr 19 at 20:32
• For $b>25$ the possible $a$ are located in the range of $\{1,\ldots,b^2\}$. Reason: Let $s(n)$ be the decimal digit sum of $n$. Then, since the number of digits of $n$ is $1+\lfloor\log_{10}n\rfloor$, you have $s(n)\le 9(1+\lfloor\log_{10}n\rfloor)\le 9(1+\log_{10}n)$. Hence, $s(a^b)\le 9(1+b\log_{10}a)$. Put $f(a) := a-9(1+b\log_{10}a)$. Then $f'(tb^2) = 1-\tfrac 9{tb\log(10)} > 0$ for $t\ge 1$ and $b>25$. Moreover, $f(b^2) = b^2-18b\log_{10}b-9 > 0$ for $b>25$. This shows that $a\le 9(1+b\log_{10}a)$ can only happen if $a\le b^2$. – amsmath Apr 19 at 20:52