# Why are some numbers disproportionately represented when calculating a digit-sum?

I wrote a small python script to calculate a number's digit-sum (i.e #152 = 1 + 5 + 2 = 8) after being raised to various powers. Then I noticed certain numbers are dramatically more common than others, while others don't occur at all.

For example, after iterating through the numbers 1-1000, each being raised to the 1-100th power, not once did a digit-sum of 3 or 6 occur, whereas 1 and 9 occur tens of thousands of times. Below are the output results;

i: 0 | Count: 0
i: 1 | Count: 27178
i: 2 | Count: 3547
i: 3 | Count: 0
i: 4 | Count: 10862
i: 5 | Count: 3546
i: 6 | Count: 0
i: 7 | Count: 10862
i: 8 | Count: 9208
i: 9 | Count: 32601


And here's a github link to the python script I wrote; https://github.com/Your-Pal-Al/Digit_Sum_Exp/blob/322846f690b8fdac9b600150de8887d7e25a734a/digit_sum_exp.py

EDIT: Fixed script loop logic errors

• @Troposphere In OP's Python script they start from powers of 2. Jul 29 at 23:51
• Well, if $k$ is a multiple of $3$ then $k^{100}$ is a multiple of $9$ so the digit sum will be $9$. If $k$ is not a multiple of $3$ then $3^{100}$ isn't either and the sum of the digits will not be $3, 6$ or $9$. So $\frac 13$ of the digits will be $9$ and $3$ and $6$ will never occur. Jul 30 at 1:29

First, your script doesn't really "iterat[e] through the numbers 1-1000, each being raised to the 1-100th power". Because you never reset the $$x$$ variable between runs of the inner loop, you're actually computing $$(a+2)^{\lfloor a/98\rfloor+2}$$ for $$0.

This creates some minor differences with the nice counts you would get if you were looping $$x$$ and $$y$$ independently from $$2$$ to $$999$$ and $$2$$ to $$99$$. Never mind, the overall pattern is broadly speaking the same.

The iterated digit sum of a positive integer is none other than the remainder when dividing the original number by $$9$$, except when the remainder is $$0$$ the digit sum is $$9$$ instead. This is the basis for casting out nines.

Now whenever $$x$$ in $$x^y$$ was divisible by $$3$$ and $$y\ge 2$$, then $$x^y$$ is divisible by $$3^2=9$$, so all those cases (about a third of all) yield the digit sum $$9$$. On the other hand, $$x^y$$ cannot be divisible by $$3$$ unless $$x$$ is divisible by $$3$$, so digit sums of $$3$$ or $$6$$ are impossible.

When $$x$$ is not divisible by $$3$$ it is coprime to $$9$$, and therefore Euler's theorem says that $$x^y\equiv 1 \pmod 9$$ whenever $$y$$ is a multiple of $$\varphi(9)=6$$. Of course, the remainder is also $$1$$ when $$x\equiv 1 \pmod 9$$ no matter what $$y$$ is. This creates a clear overweight of $$1$$ among the sums.

Finally, a square modulo $$9$$ is always one of $$\{0,1,4,7\}$$ -- and $$x^y$$ is a square whenever $$y$$ is even. This explains why you see $$4$$ and $$7$$ more often than $$2$$, $$5$$, or $$8$$. On the other hand, a cube modulo $$9$$ is always one of $$\{0,1,8\}$$, so when $$y\equiv 3\pmod 6$$, a third of the $$x$$ values produce $$8$$, explaining why $$8$$ is more popular than $$2$$ and $$5$$.

We can also count the popularity of each residue directly by writing up a table: $$\begin{array}{r|cccccc} y = & 6n+6 & 6n+7 & 6n+2 & 6n+3 & 6n+4& 6n+5 \\ \hline 0^y \bmod 9 = & 0 & 0 & 0 & 0 & 0 & 0 \\ 1^y \bmod 9 = & 1 & 1 & 1 & 1 & 1 & 1 \\ 2^y \bmod 9 = & 1 & 2 & 4 & 8 & 7 & 5 \\ 3^y \bmod 9 = & 0 & 0 & 0 & 0 & 0 & 0 \\ 4^y \bmod 9 = & 1 & 4 & 7 & 1 & 4 & 7 \\ 5^y \bmod 9 = & 1 & 5 & 7 & 8 & 4 & 2 \\ 6^y \bmod 9 = & 0 & 0 & 0 & 0 & 0 & 0 \\ 7^y \bmod 9 = & 1 & 7 & 4 & 1 & 7 & 4 \\ 8^y \bmod 9 = & 1 & 8 & 1 & 8 & 1 & 8 \end{array}$$

Counting instances of each result in this table you should match up your experimental frequencies pretty well. (Each of the cells is hit by about $$\frac{98\cdot 998}{54}\approx 1811$$ of your test cases).