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Well I am quite sure it's known (I mean number theory exists thousands of years), warning beforehand, it may look like numerology, but I try not to go to mysticism.

So I was in a bus, and from boredom I started just adding numbers in the next way: $$1+1=2$$ $$2+2=4$$ $$4+4=8....$$

etc up to $32,768$ (it was quite boring, I can tell... :-)), I didn't have calculator.

and notice that if I keep adding the digits until I get a number from 1 to 10, I get that for example for $8+8=16$, $1+6=7$, now seven steps after this, at $512+512=1024$, which $1+2+4=7$, and again after $7$ steps $32768+32768=65536$, and adding $6+5+5+3+6=10+12+3=25$, $2+5=7$. So this led me to cojecture that this repetition may occur endlessly.

Now ofcourse I can programme some code that will check for large numbers, but I am tired, long day. So if this indeed the case (which could be disproved, but even then I would wonder when this repetition stops) then why?

As I said, I am tired, it may make no sense, and I might have done mistakes in my calculations, and it may be trivial.

Either way, if you have some answer, I would like to hear it.

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Interesting! I've never heard of such a pattern before so it may very well be original. I'd love to hear what happens when you write that computer program. – Samuel Reid Dec 29 '11 at 18:33
As a layman, the sheer number of patterns that mathematicians have already found and documented amazes me. I suppose math's been around for a long while, but still! – Reid Dec 30 '11 at 6:26
up vote 42 down vote accepted

"Adding the digits until [you] get a number from $1$ to $10$" is the same as finding the remainder when dividing by $9$ (except that you would get $9$ instead of $0$ by adding digits, and there is no reason to stop with $10$: you can add the digits to get $1$).

It turns out, by something called Euler's Theorem, that if $a$ is not divisible by $3$, then $a^6$ always leaves a remainder of $1$ when divided by $9$. In particular, $2^6$ has remainder $1$ when divided by $7$.

Another property of remainders is that if $a$ leaves a remainder of $r$ when divided by $9$, and $b$ leaves a remainder of $s$, then $ab$ leaves the same remainder as $rs$.

So, that means that if $2^n$ leaves a remainder of $r$, then $2^n\times 2^6 = 2^{n+6}$ will leave the same remainder as $r\times 1 = r$; that is, the same remainder as $2^n$. So adding the digits of $2^n$ until you get a single number between $1$ and $9$ will give you the same answer as doing it for $2^{n+6}$. This is what you observed: $8=2^3$, and $512=2^9 = 2^{3+6}$. You will get the same answer ($7$) with $2^{15}$, $2^{21}$, $2^{27}$, etc.

I note that you were slightly off in describing $512$ as being "seven steps after" $8$: it's really only six steps later: $$8\stackrel{1}{\mapsto} 16 \stackrel{2}{\mapsto}32 \stackrel{3}{\mapsto}64\stackrel{4}{\mapsto}128\stackrel{5}{\mapsto}256\stackrel{6}{\mapsto}512.$$

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OK, thanks. I should have known it was from something I learned already, I thought to myself that this must be true but I guess that I forgot from this theorem. – MathematicalPhysicist Dec 30 '11 at 3:43
Also, $2^6 = 64$ happens to have the same digital root as its reciprocal $2^{-6} = \frac{1}{64} = 0.015625$, namely $1$. Therefore, for any given $n \in \mathbb{Z}$, all of the numbers $\{ 64^k \cdot n \ | \ k \in \mathbb{Z}\}$ have the same digital root as $n$. – r.e.s. Dec 30 '11 at 4:20

This is indeed the case.

The value you get out at the end of all your summations is just the value of your number mod 9 (this is because any natural number is congruent to the sum of its digits mod 9). For example, $1024=9\cdot113+7$ is congruent to 7 mod 9. Also note that the numbers you're getting out by your procedure of successive doubling are just the powers of 2. Finally, I think you're off by one in your counting of sevens -- the powers of two that are working for you are $2^4=16$, $2^{10}=1024$, $2^{16}=65536$, etc.

So your fact is true because it's true for $2^4$, and each subsequent term differs by a factor of $2^6$ which is congruent to 1 mod 9 (either by direct computation or by Euler's Theorem -- $\phi(9)=6$).

So in general, the sum of the sum of the .... sum of the digits of your values $2^{4+6k}$ is given by $$ 2^{4+6k}\equiv 2^4\cdot (2^6)^k\equiv 7\cdot 1^k\equiv \boxed{7}\pmod{9}. $$

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There is indeed a period of $6$. The sums of digits cycles through the numbers $1,2,4,8,7,5$.

You are doubling the number each time, but then the sum of the digits of that number will double as well, since it is a linear function of the number. This then shows that if you start with $1$ you will go through the cycle

$$1 \to 2 \to 4 \to 8 \to 16=7 \to 14=5 \to 10=1 \to \ldots$$

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"You are doubling the number each time, but then the sum of the digits of that number will double as well, since it is a linear function of the number." -- This claim is strange, and not quite correct if interpreted literally; e.g., the sum of digits in $55$ is $10$ whereas it's $2$ for $110$. Are you doubling each digit independently and adding? – Srivatsan Dec 29 '11 at 20:25
And $1+0=1$. Therefore, doubled. – Raskolnikov Dec 30 '11 at 8:42

A repetition of this sort was bound to happen, and it always happens even under more general circumstances.

First, as others have pointed out, the sum of the base-10 digits of a number $N$ is congruent to $N$ modulo $9$. The reason for this is that $10\equiv 1 \bmod 9$, and so $$\begin{align*} N &= a_t \cdot 10^t + a_{t-1} \cdot 10^{t-1} +\cdots +a_2 \cdot 10^2 + a_1 \cdot 10 +a_0 \\ & \equiv a_t\cdot 1^t + a_{t-1}\cdot 1^{t-1} +\cdots +a_2\cdot 1^2 + a_1 \cdot 1 +a_0 \bmod 9\\ & \equiv a_t + a_{t-1} +\cdots +a_2 + a_1 +a_0 \bmod 9. \end{align*}$$ Thus, if you add all the digits of $N$ and obtain $N_1$, then $N\equiv N_1\bmod 9$. If now we add all the digits of $N_1$ and obtain $N_2$, then $N\equiv N_1\equiv N_2 \bmod 9$. In this way we may create a sequence $N>N_1>N_2> \cdots$, and since all $N_i$ are natural numbers, we end up at some $1\leq N_t \leq 9$, such that $N_t\equiv N\bmod 9$, so $N_t$ is simply the remainder of division of $N$ by $9$.

Let $a>1$ be any natural number relatively prime to $3$, whose sum of base-10 digits is $b$, and let $s$ be the order of $b\bmod 9$, i.e., $s$ is the smallest positive number such that $b^s\equiv 1 \bmod 9$. Then:

  • The sum of the base-10 digits of $a$ is $b\bmod 9$.
  • The sum of the base-10 digits of $a^{1+sk}$ is also $b\bmod 9$, for all $k\geq 0$, because $$a^{1+sk}\equiv b^{1+sk}\equiv b\cdot b^{sk}\equiv b\cdot (b^s)^k\equiv b \cdot 1\equiv b \bmod 9.$$
  • For a fixed $t\geq 1$, the sum of the base-10 digits of $a^{t+sk}$ is $b^t\bmod 9$, for all $k\geq 0$, for similar reasons as above.
  • For a fixed $t\geq 1$, the sum of the base-10 digits of $$a^{t+sk} + a^{t+sk}$$ is $2\cdot b^t \bmod 9$, for all $k\geq 0$.
  • And more generally, for fixed $r\geq 1$ (relatively prime to $3$) and $t\geq 1$, the sum of the base-10 digits of $$a^{t+sk} + \cdots + a^{t+sk} = r\cdot a^{t+sk},$$ where we have added $r$ copies of $a^{t+sk}$, is $r\cdot b^t \bmod 9$, for all $k\geq 0$.

Your example is the case where $a=2$, $b=2$, $s=6$, $t=3$ and $r=2$. According to the formula above, the sum of the digits must be $$r\cdot b^t \equiv 2\cdot 2^3\equiv 7 \bmod 9.$$ But any other choice works just as well. Pick $a=11$, $b\equiv a\equiv 2 \bmod 9$, $s=6$, $t=2$ and $r=5$. Then, the sum of the digits of the numbers $$11^{2+6k}+11^{2+6k}+11^{2+6k}+11^{2+6k}+11^{2+6k},$$ for all $k\geq 0$, is congruent to $$r\cdot b^t \equiv 5\cdot 2^2\equiv 2 \bmod 9.$$ For instance:

  • $11^{2}+11^{2}+11^{2}+11^{2}+11^{2}=605,$ and $6+5=11$ and $1+1=2$.

  • $11^{8}+11^{8}+11^{8}+11^{8}+11^{8}=1071794405,$ and $$1+0+7+1+7+9+4+4+5=38$$ and $3+8=11$, and $1+1=2$.

  • Etc.

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