Japan Junior Mathematical Olympiad

Let us show first that any positive integer $n$ satisfying the condition of the problem must have the same number in all of its digits. In order to show this it is enough to show that if the integers $a$ and $b$ appear among the digits of $n$, where $1\le a,b\le 7$ and if $n$ satisfies the condition of the problem, then $a=b$ must hold.

So, suppose $a$ and $b$ appear among the digits of the number $n$. Permute the digits of $n$ in such a way to get a number in which $a$ appears in the 10's digit and $b$ appears in 1's digit. Call this new number $n^{\prime }$. Let $N$ be the number obtained by discarding the 10's and 1's digits of $n^{\prime }$ (In case $n^{\prime }$ is a 2-digit number, let $N=0$). Then we can write $n^{\prime }=100N+10a+b$. Let $n^{\prime \prime }$ be the number obtained by interchanging the 10's and 1's digits of $n^{\prime }$. Then we also have $n^{\prime \prime }=100N+10b+a$.

By the condition of the problem both $n^{\prime }$ and $n^{\prime \prime }$ are the multiples of $7$ since both are obtained by permuting digits of $n$. Consequently, the difference $n^{\prime }-n^{\prime \prime }=(100N+10a+b)-(100N+10b+a)=9(a-b)$ must also be a multiple of $7$. Since $7$ and $9$ are relatively prime, we conclude that $a-b$ must be a multiple of $7$. But since $1\le a,b\le 7$, we conclude that $a=b$, which establishes our claim.

Next, consider the case where $1\le a\le 6$. By carrying out simple divisions, we see that the 6-digit number $111111$ is a multiple of $7$, while any $j$-digit number obtained by putting $1$ into each digit is not a multiple of $7$ if $1\le j\le 5$.

Now, for each positive integer $k\ge 1$ and $\ell$ with $0\le \ell \le 5$, denote by $N(6k+\ell )$ the number having $6k+\ell$ digits, each of which is $a$. Then if we divide $N(6k+\ell )$ by $111111$ we get as a remainder a number $m_{\ell }$ which has $\ell$ digits each of which is $a$ (We let $m_0=0$, since there is no remainder if $\ell =0$). Consequently, we see that in order to decide whether a number $N(6k+\ell )$ is a multiple of $7$ or not it is enough to see whether the remainder $m_{\ell }$ is a multiple of $7$ or not.

If $\ell =0$, $N(6k)$ is a multiple of $7$. So, let us suppose that $1\le \ell \le 5$. Then, $m_{\ell }$ is a multiple by $a$ of the number $t$ having $\ell$-digits, each of which is $1$. Since $1\le \ell \le 5$, we see that $t$ is not multiple of $7$. Since $a$ is also not a multiple of $7$, we conclude that $m_{\ell }$ is not multiple of $7$ either. Thus, we see that the number with all of its digits being equal to $a$, where $1\le a\le 6$, must be of the form $N(6k)$, where $k\ge 1$.

Summarizing the arguments made above, we conclude that the positive integer $n$ satisfying the condition of the problem must have one of the following forms:

$$\bullet n=\underset{6k\text{ times}}{\underbrace{aa\cdots aa}}(1\le a\le 6,k\ge 1),$$ $$\bullet n=\underset{p\text{ times}}{\underbrace{77\cdots 77}}(p\text{ is an arbitrary positive integer}).$$