Proving 111...111 is divisible by $3^n$

Question

Problem: Show that the number $11...11$ with $3^n$ digits is divisible by $3^n$.

My Attempt I tried solving this problem on base 3. It is easy to see that

$$3^n = 100....00$$ (upto $n$ 0's)

And $$ 11...11 = a_0 + 3a_1 + 3^2a_2 + .... 3^ma_m $$ if the the claim is correct then that would would imply that $11..11$ in base 3 would end with $3^n$ zeroes then that would mean that, $$ a_1=a_2=a_3...=a_k=0$$where k= $3^n$ I tried proving this but wasn't able to, subsequesntly I also tried to prove that a number is divisible by $3^n$ if it's sum is a multiple of $3^n$. Any insight on solving this?

(PS I do know this already has been posted on MSE but I wanted to see if I could prove this using my method)

I do not yet know how to use induction but I will try proving this once again when study it. — bm27, Jun 16 '23 at 17:48
Does this answer your question? Is the repunit $R_{3^k}$ divisible by $3^k$? — Martin R, Jun 16 '23 at 17:55
@MartinR it does answer the problem but I was looking for insights into my method, as to how to go forward — bm27, Jun 16 '23 at 18:00
@Calvin Lin: I saw bad the punctuation in the calculator. Thanks you very much. — Ataulfo, Jun 16 '23 at 18:16
@bm27 The crux is that $R_{k+1} = R_k \times 1000...001000...001$, so there is an additional power of 3. $\quad$ FWIW Your method doesn't introduce a new idea that provides insight. We could rewrite the solution in your terminology to show that $a_{k+1} = 0 $, which is entirely based on the factoring. — Calvin Lin, Jun 16 '23 at 19:05
The crux doesn't require induction. It's just an algebraic statement/form. $\quad$ Using the crux to prove the statement requires induction (in some manner). — Calvin Lin, Jun 18 '23 at 06:20

score 2 · Answer 1 · edited Jul 06 '23 at 14:23

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Here is one possible way of doing it using some high school math competition technique: $$ 111....111 \quad \text{(with $3^n\ 1s)$} = \sum_{i=0}^{3^n-1} 10^i = \frac{1}{9} (10^{3^n} -1 ). $$

We count number of factor 3 in the numerator by lifting the exponent: https://en.wikipedia.org/wiki/Lifting-the-exponent_lemma

$$ \nu_3(10^{3^n} -1) = \nu_3(10-1) + \nu_3(3^n) = n+2 $$

Canceling out the 9 in the denominator which contains two 3's, we get $$ \nu_3(111....111) = n $$

edited Jul 06 '23 at 14:23

J. W. Tanner

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answered Jun 16 '23 at 18:12

minqiang

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Without resorting to LTE, we could use the telescoping product $\prod (10^{3^{k+1} } - 1 ) / (10^{3^k} - 1 ) $ to explain where the powers of 3 come from. This is the induction proof in the linked question. – Calvin Lin Jun 16 '23 at 19:02

CopyPasteIt · Answer 2 · 2023-06-19T18:40:30.200

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If $a$ is divisible by $3^k$ for fixed $k \ge 0$ and

$$ s \equiv 1 \pmod{3} \;\land\; t \equiv 1 \pmod{3} \;\land\; u \equiv 1 \pmod{3}$$

then by algebra

$$ 3^{k+1} \mid sa + ta + ua$$

Observe that $10^n \equiv 1 \pmod{3}$ for $n \ge 0$.

Using the above we'll make some non-abstract statements that the OP can formalize into a rigorous proof using induction:

$\; 3^1$ divides $111$
$\; 3^2$ divides $10^6 \cdot 111 + 10^3 \cdot 111 + 111$
$\; 3^2$ divides $111111111$

edited Jun 19 '23 at 18:40

answered Jun 19 '23 at 10:00

CopyPasteIt

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(Responding to your deleted question, which you might already have realized) The denominator is $g-1$. Thankfully, you set $g -1 > 0 $. $\quad$ When $p$ is not a prime, even though $g^n - 1 \equiv 0$, we have to be careful when dividing out by $g-1$ esp if $ \gcd (g-1, p ) \neq 1$. There is a bit more work to be done to show that there is some sum (basically, take enough multiples of $n$ to ensure the denominator cancels out. EG $g-1$ is more than sufficient. – Calvin Lin Jun 22 '23 at 23:34

Proving 111...111 is divisible by $3^n$

2 Answers2

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