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Prove that $\left(\forall x \in \mathbb{N}\right)\left(\exists y \in \>\mathbb{N}\right)\left(\underbrace{3...3}_\text{x}|\underbrace{1...1}_\text{y}\right)$ and find the smallest $y$ for each $x$.

Note: "$a|b$" means "$a$ divides $b$".

I know that for $x \in \mathbb{N}, k \in \mathbb{Z}:$

$$3|\underbrace{111}_\text{3}, 3|\underbrace{111111}_\text{6}, 3|\underbrace{1...1}_\text{9}, 3|\underbrace{1...1}_\text{3k}; \\ 33|\underbrace{111111}_\text{6}, 33|\underbrace{1...1}_\text{12}, 33|\underbrace{1...1}_\text{$3 \cdot 2 \cdot k$}; \\ \underbrace{3...3}_\text{x}|\underbrace{1...1}_\text{$3 \cdot x \cdot k$}.$$

So $y=3xk$ satisfies the conditions and the smallest $y$ equals $3x$, but how can i prove this formally?

My attempt at solution:

Edit:
I wanted to prove that $y$ must be $3lx$ and so solve both problems. But now I realized that my method does not work. Now I want to understand how to prove that $3x$ is the smallest possible y if I prove (by induction or by any other method) that there always exists $k$ for $y = 3lx$.

$$\left(\forall x \in \mathbb{N}\right)\left(\exists y \in \mathbb{N}\right)\left(\underbrace{3...3}_\text{x}|\underbrace{1...1}_\text{y}\right) \\ \iff \\ \left(\forall x \in \mathbb{N}\right)\left(\exists y \in \mathbb{N}\right)\left(\left(\exists k \in \mathbb{Z}\right)\left(\sum_{i=0}^{y-1} 10^i = 3 \cdot \sum_{i=0}^{x-1} 10^i \cdot k\right)\right)$$ By the definition of repunit (number that contains only the digit 1):$$ \sum_{i=0}^{n-1} 10^i = \dfrac{10^n-1}{9}, n \geq 1,$$ then $$\dfrac{10^y-1}{9} = 3k\cdot\dfrac{10^x-1}{9}.$$

So $\dfrac{10^y-1}{9}$ is divisible by 3 and $\dfrac{10^x-1}{9}$.

For 3:

$$\dfrac{10^y-1}{9} = 3l \\ \underbrace{1...1}_\text{$y$} = 3l \\ \underbrace{1 +\ ...\ + 1}_\text{$y$} = 3l\ (*) \\ y = 3l$$

(*): Divisibility rule for 3.

For $\dfrac{10^x-1}{9}$:

$$\dfrac{10^y-1}{9} = \dfrac{10^x-1}{9} \cdot m \\ y = xm\ (**)$$

(**): Proof: Prove that if $\frac {10^n-1}{9} | \frac {10^m-1}{9}$, then $n|m$.

So $y = lcm(3, x) = \dfrac{3x}{gcd(3,x)}$ and i stuck there. Any ideas please?

Thanks in advance.

Dia
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  • My first instinct before reading "and find the smallest y for each x" was to just say that $\frac{111}{3}\cdot 3=111,\frac{10101}{3}\cdot 33=111111, \frac{1001001}{3}\cdot333=111111111,\dots$ and in general $(1\underbrace{00\dots 0}{x-1}1\underbrace{00\dots0}{x-1}1\cdot \frac{1}{3})\cdot \underbrace{33\dots 3}_x=\underbrace{11\dots 1}_x\underbrace{11\dots 1}_x\underbrace{11\dots 1}_x$. If you formalize my argument and observation (and prove it is correct) then that takes care of the existence part of the proof. I have not checked if this is also the smallest. It might very well be. – JMoravitz Nov 13 '17 at 17:30
  • Do you want to know the reason of condition for divisibility of numbers of form $111...111$ by 3 ? – sirous Nov 13 '17 at 19:39
  • Just glancing. $lcm(3,x)$ is either $x$ if $3|x$ or $3x$ if $3\not \mid x$. Because $3$ is prime.But I didn't actually read enough to see why you wanted that. – fleablood Nov 13 '17 at 20:03
  • In reading it really sounds like you are starting with a conclusion and trying to prove it implies something. You should try to avoid starting with a conclusion. But if you do, be absolutely clear that we you get as a result is completely equivalent (every step must be an if and only if statement.) – fleablood Nov 13 '17 at 20:07
  • I wanted to prove that y must be 3lx and so solve both problems. But now I realized that my method does not work. Now I want to understand how to prove that 3x is the smallest possible y if I prove (by induction or by any other method) that there always exists k for y = 3lx. – Dia Nov 13 '17 at 23:14

3 Answers3

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You more or less have it.

Let $333......33$ $k$ threes and $11111.....1$ $k$ times be written as $3_k$ and $1_k$ respectively.

$3_x = 3*1_x|1_y$ will mean $3|1_y$ $1_y$ has a multiple of $3$ $1$s. It will also mean $1_x|1_y$ which, as you researched, means $x|y$.

So if $3\not \mid y$ then $x|y \implies 3\not \mid x$ we need $y$ to be the smallest multiple of both $x$ and $3$. That is ... $y = 3x$.

But what if $3|x$? And so $x|y$ and $3|y$. Say $x=3^mk$ and $\gcd(3,k) = 1$ Then $1_x = 1_k*100....10000....1:= 1_k*1_{3^m,k}$ where $1_{a,b} =$ $a$ $1$ separated by $b-1$ $0$s (or other words $\sum_{j=0}^{a} 10^{j*b}$).

So $3_x|1_y$ means $3|1_y$ and $1_x|1_y$ and $1_k|1_y$ and $1_{3^m,k}|1_y$. Let $V = \frac {1_y}{1_3,k}$ which must be equal to $1_j$ where $k|j$. But then $3*1_k|V$ so $V$ is divisible by $3$ and $3|j$. So $y$ must be a multiple of $3x$.

So the smallest such multiple of $x$ satisfiying all that is still $3x$.

fleablood
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  • I thought that my method was wrong, but you managed to complete my proof! Thank you for your help. I appreciate it very much! – Dia Nov 15 '17 at 02:21
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The existence proof is easy. Suppose there are $n$ 3's in your proposed divisor $d$. Form the number $M$ with $3n$ 1's. Then since the number of 1's in $M$ is divisible by $n$, $M$ will be divisible by $d/3$ (a number with $n$ 1's). By the digital sum rules for divisibility by powers of $3$ (for divisibility by $3^k$ with $k\ge 2$, blocks of $3^{k-2}$ digits must add up to a multiple of $3^k$), tripling the number of 1's in $M$ from $n$ to $3n$ will then provide an additional factor of $3$ that completes the divisibility requirement.

Example: $n=3, d=333$ gives $111,111,111$ as the candidate. We then have $111,111,111=1,001,001×111=333,667×333$.

This does NOT, however prove that smallest solution.

Oscar Lanzi
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  • Thank you! Do you have any ideas how to prove that 3n is the smallest solution? – Dia Nov 13 '17 at 23:18
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$33...33 = 3 ˟ 11...11 $

if number of digits of $11...11 $ is $2p$,(p=1,2,3...) then it is divisible by $11...11$ with number of digits $2q≤ p$, i.e. we must have:

$n_1=2p$

Also $11...11$ is divisible by $3$ if the sum of digits is divisible by 3. That means we must have:

$n_1 = 3 k$

That is if the number of digits of $11...11$ is $n_1=6pk$ or generally $n_1=6 t$ (t any natural number) then a number like $33...33$ can divide it if the number of its digits is $n_3=6s, n_3≤ n_1/2$

sirous
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