# Prove that every number ending in a $3$ has a multiple which consists only of ones.

Prove that every number ending in a $3$ has a multiple which consists only of ones.

Eg. $3$ has $111$, $13$ has $111111$.

Also, is their any direct way (without repetitive multiplication and checking) of obtaining such multiple of any given number( ending with $3$) ?

• Either answer to this related question solves this easily. – Chris Eagle Jun 27 '12 at 14:12
• @ChrisEagle: that question allows digits $0$, this one doesn't. – Marc van Leeuwen Jun 27 '12 at 14:13
• @Marc: Yes, I know. However, my answer covers this case explicitly, while N. S.'s answer is easily extended to this case. – Chris Eagle Jun 27 '12 at 14:15
• @MarcvanLeeuwen After arranging a multiple which has all ones followed by all zeros, you may divide through by 10 as many times as you like to remove the zeros (since a number ending in 3 is coprime with 2 and 5). – rschwieb Jun 27 '12 at 14:27
• Off topic, but I think this prime factorization is a little neat. Doubtless there are more amazing ones but: $11111111111=21649\cdot 513239$ – rschwieb Jun 27 '12 at 18:22

The number $111...11$ with $n+1$ digits is $1 + 10 + ... + 10^n = \frac{10^{n+1} - 1}{9}$ using the formula for geometric progressions.

Now any number $x$ which ends in three is coprime to $10$, and so is $9x$: $10$ is a unit $\mod 9x$. In particular, you have the Euler-Fermat theorem: $10^{\varphi(9x)} \equiv 1 \mod 9x$. This means that $9x$ divides $10^{\varphi(9x)} - 1$, so $x$ divides $\frac{10^{\varphi(9x)} - 1}{9}$.

So $x$ divides the number $111..11$ with $\varphi(9x)$ digits, where $\varphi$ is Euler's phi function.

If $n$ ends with a $3$ it must be coprime to $10$. Then, by Fermat's little theorem, $$10^{\varphi(9n)} \equiv 1 \pmod {9n},$$ so $n$ divides $(10^{\varphi(9n)}-1)/9$ whose decimal representation consists only of ones.

We can prove more very simply: if $$\rm\,n\,$$ is coprime to $$10\,$$ then any number with $$\rm\, n\,$$ digits all $$\ne 0$$ has a contiguous digit subsequence that forms a number divisible by $$\rm\,n.\,$$ Suppose the digits are $$\rm\,d_{n}\ldots d_1.\,$$ By $$\rm\,d_i\ne 0\,$$ the $$\rm\,n\!+\!1\,$$ numbers $$\rm\,0,\,d_1,\, d_2 d_1,\, d_3 d_2 d_1,\, \ldots,d_n\!\ldots d_1$$ are distinct. By Pigeonhole $$\rm\color{#c00}{two\ are\ congruent}$$ $$\rm\,mod\ n,\,$$ so $$\rm\,n\,$$ divides their difference $$\rm = 10^k\,$$ times the number $$\rm\,\color{#0a0}{e\ne 0}\,$$ formed by the $$\rm\color{#0a0}{extra\ digits}$$ of the longest, so $$\rm\,n\,|\,10^ke\,$$ $$\Rightarrow$$ $$\rm\,n\mid e,\,$$ by $$\rm\,n\,$$ is coprime to $$10$$ and Euclid's Lemma.

Let's do a simple example: $$\rm\,n=9,\,$$ with number $$\rm\,98765\color{#0a0}{432}1.\,$$ Initial digit substrings $$\!\bmod 9\,$$ are
\begin{align}\bmod 9\!:\quad\ \ \color{#C00}{1}&\equiv\color{#c00}1\\ 21&\equiv 3\\ 321&\equiv 6\\ \color{#C00}{4321}&\equiv\color{#c00} 1\end{align}\qquad

therefore we conclude that $$\rm\,9\mid\color{#c00}{4321\!-\!1} = \color{#0a0}{432}\cdot 10,\,$$ so $$\,9\,|\,\color{#0a0a}{432}.$$

In OP the divisor $$\rm\,n=10k+3\,$$ is coprime to $$10$$, so the number $$\,11\ldots 11$$ $$\rm\,(n$$ digits) does the trick, i.e. some digit subsequence $$\color{#0a0}{11\ldots 11}$$ forms an integer divisible by $$\rm\,n.$$

The result extends to any number having $$\rm\,n\,$$ nonzero digits: simply take the subsequences beginning with nonzero leading digit. This implies that the $$\rm\,n+1\,$$ numbers are increasing (so distinct), and the number formed by the extra digits is nonzero, since its leading digit is nonzero.

Call your number $n$. Since $9n$ is relatively prime with $10$, the number $10$ is invertible modulo $9n$, so there is some power $10^k$ (with in fact $k$ a divisor of $\phi(9n)$) that is congruent to $1$ modulo $9n$, in other words $9n$ divides $10^k-1$, a number formed of $k$ digits $9$. But then $n$ divides the number with $k$ digits $1$.

Find a more generalized solution for any natural number in base here: All odd primes except $5$ divide a number made up of all $1$s