I was messing around and found something interesting, at least to me. Although I may not know how to delicately word this, so I hope it is clear.

Claim: Each natural number $n$, whose sequence of digits are represented by the string $S$ is divisible by $3$ if the length of $S$ is divisible by $3$, and each consecutive sequence of $3$ characters in $S$ are identical numbers.

Probably this idea can be pinned down even further. But even with what I have added, the part where it can be longer than $3$ digits being the same number, it seems still challenging at least for me to prove. Maybe if we first prove that a $3$ digit number is divisible by $3$ if each digit is the same then it will be easier? Or perhaps Divisibility by Liljevalch’s theorem is better yet. But in any case we can say that the three consecutive digits sum to $3m$ (where $m$ is the identical digit), which is divisible by $3$.

How might you go about something like this. But probably more important, how do I first word this theorem better since I am not yet literate with such things? Perhaps without needing to talk about strings?


1 Answer 1


Theorem. Let $N$ be a natural number whose decimal expansion is of the form: $$N= a_0(1+10+10^2)+a_1(10^3+10^4+10^6)+\cdots+a_{n}(10^{3n}+10^{3n+1}+10^{3n+2}),$$ for some $n\geq 0$ and $a_i\in\{0,1,2,\ldots,9\}$ for each $i=0,\ldots,n$. Then $N$ is divisible by $3$.

Proof. Let us reduce $N\bmod 3$. We obtain:

\begin{align*} N &\equiv a_0(1+10+10^2)+a_1(10^3+10^4+10^6)+\cdots+a_{n}(10^{3n}+10^{3n+1}+10^{3n+2})\\ & \equiv a_0(1+1+1)+a_1(1+1+1)+\cdots+a_n(1+1+1)\\ &\equiv a_0\cdot 0 + a_1\cdot 0 + \cdots + a_n\cdot 0 \\ & \equiv 0 \bmod 3. \end{align*} Therefore, $N\equiv 0\bmod 3$ and this is equivalent to $N$ being divisible by $3$.

  • 1
    $\begingroup$ Beat me to it, and exactly right. +1 $\endgroup$
    – davidlowryduda
    Commented Mar 12, 2013 at 2:23
  • $\begingroup$ The contrast in the clarity between my claim and your theorem is palpable, well done sir. $\endgroup$
    – Leonardo
    Commented Mar 12, 2013 at 2:25
  • 1
    $\begingroup$ Notice also that once you have the correct statement, the proof is trivial! $\endgroup$ Commented Mar 12, 2013 at 2:27

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