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For which integers $k$ is it possible to have $b=ka$, where $a$ is a positive integer and $b$ is obtained from $a$ by moving the initial digit of $a$ to the end.

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    The trivial answer would be "k=1" and "b" and "a" are numbers like "11" "222" "333" "44444" etc – 2017-02-21
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    Are you using base ten? – 2017-02-21
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    $k=5$ is impossible, as I showed [here](http://math.stackexchange.com/questions/2153892/prove-that-it-is-impossible-to-have-integers-b-5a-under-a-digit-re-ordering/2153899#2153899). – 2017-02-21
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    $k=3$ works, for instance with $a=142857$. – 2017-02-21
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    Not possible for $k\ge 5$. The closest we get to $k=5$ is if $a$ starts with a $1$ and the second digit of $a$ is a $9$. Other starting digits give lower max values of $k$. – 2017-02-21

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Write $a=10^n u + w$, where $u$ is a digit and $0 \le w < 10^n$. Then $b=10w+u$.

Now, $b=ka$ iff $k(10^n u + w) = 10 w +u $ iff $ (k10^n-1)u= (10-k)w$.

Since $w < 10^n$, we have $(k10^n-1)u < (10-k)10^n$ and so $(ku+k-10)10^n < u < 10$. This can only happen if $ku+k-10 \le 0$. In particular, since $u\ge 1$, this can only happen when $k\le 5$.

$k=1$ implies $w=(11\cdots1)u$.

$k=2$ implies $(2\cdot 10^n-1)u=8w$. Since $2\cdot 10^n-1$ is odd, $8$ must divide $u$ and so $u=8$. But $ku+k \le 10$ implies $u \le 4$, so no solution.

$k=3$ implies $(3\cdot 10^n-1)u=7w$ and so $7$ divides $3\cdot 10^n-1$. This implies $n \equiv 5 \bmod 6$. Also, $ku+k \le 10$ implies $u \le 2$. A sample solution is $w=42857$, which gives $a=142857$ for $u=1$ and $a=285714$ for $u=2$.

$k=4$ implies $(4\cdot 10^n-1)u=6w$. Since $4\cdot 10^n-1$ is odd, $2$ must divide $u$. But $ku+k \le 10$ implies $u \le 1$, so no solution.

$k=5$ implies $(5\cdot 10^n-1)u=5w$. So $5$ must divide $u$. But $ku+k \le 10$ implies $u \le 1$, so no solution.

Bottom line: The only solutions are $a=(11\cdots1)u$ for $k=1$ and $a$ formed by repeating $142857$ or $285714$ for $k=3$.

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    @SarahTerrep, which part of the argument are you having trouble with? – 2017-02-25
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    @SarahTerrep Note that you have $ku+k-10 \le 0 \iff ku+k \le 10$, so for $k=3$ this give $u\in \{1,2\}$ and thus $7 \nmid u$. $\quad 7 \mid (3\cdot10^n-1)$ means that $3\cdot10^n \equiv 1 \bmod 7$. Since $10\equiv 3$ we have $3^{n+1}\equiv 1 \bmod 7$ and since the multiplicative order of $3 \bmod 7$ is $6$, we know $6 \mid (n+1)$. – 2017-03-14