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

Question

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 $3$) ?

Answer 1

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.

Answer 2

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.

Answer 3

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 $1$.

Answer 4

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

Let's do a simple example: $\,n=9,\,$ with number $\,98765\color{#0a0}{432}1.\,$ 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 $\,9\mid\color{#c00}{4321\!-\!1} = \color{#0a0}{432}\cdot 10,\,$ so $\,9\,|\,\color{#0a0a}{432}.$

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

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

Answer 5

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