Problem 1042

\section*{Solution}

Suppose the digits are $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}$ and that $a_{1} a_{2} a_{3} a_{4} a_{5} a_{6}$ is divisible by 37 . We claim that $a_{2} a_{3} a_{4} a_{5} a_{6} a_{1}$ is also divisible by 37. Put $n=a_{2} a_{3} a_{4} a_{5} a_{6} a$ and $m=a_{1}$. The original number is $10^{5} \mathrm{~m}+\mathrm{n}$ and the derived number is $10 \mathrm{n}+\mathrm{m}$. But 37 divides $10^{3}-1$ and hence also $10^{6}-1$ and $\mathrm{m}\left(10^{6}-1\right)$. So it also divides $10\left(10^{5} \mathrm{~m}+\mathrm{n}\right)-\mathrm{m}\left(10^{6}-1\right)=10 \mathrm{n}+\mathrm{m}$, which proves the claim

Iterating, we get the original number and 5 others:

```

a
a}2\mp@subsup{a}{4}{}\mp@subsup{a}{5}{}\mp@subsup{a}{6}{}\mp@subsup{a}{1}{
a3a
a
a}\mp@subsup{}{6}{*
a

```

Similarly, we have that $a_{1} a_{2} a_{3} a_{4} a_{5} a_{6}-a_{1} a_{2} a_{6} a_{4} a_{5} a_{3}=\left(a_{6}-a_{3}\right) 999$, so $a_{1} a_{2} a_{6} a_{4} a_{5} a_{3}$ is also divisible by 37 . Iterating we get:

$a_{1} a_{2} a_{6} a_{4} a_{5} a_{3}$

$a_{2} a_{6} a_{4} a_{5} a_{3} a_{1}$

$a_{6} a_{4} a_{5} a_{3} a_{1} a_{2}$

$a_{4} a_{5} a_{3} a_{1} a_{2} a_{6}$

$a_{5} a_{3} a_{1} a_{2} a_{6} a_{4}$

$a_{3} a_{1} a_{2} a_{6} a_{4} a_{5}$

Similarly, we could swap the first term and the fourth to get:

```

a}4\mp@subsup{}{2}{* a
a
a3a
a
a
a

```

or the second and the fifth to get:

$a_{1} a_{5} a_{3} a_{4} a_{2} a_{6}$

$a_{5} a_{3} a_{4} a_{2} a_{6} a_{1}$

$a_{3} a_{4} a_{2} a_{6} a_{1} a_{5}$

$a_{4} a_{2} a_{6} a_{1} a_{5} a_{3}$

$a_{2} a_{6} a_{1} a_{5} a_{3} a_{4}$

$a_{6} a_{1} a_{5} a_{3} a_{4} a_{2}$