$\overline {abc}$ is divisible by $37$. Prove that $\overline {bca}$ and $\overline {cab}$ are also divisible by $37$.
$$\overline {abc} = 100a + 10b + c$$
$$\overline {bca} = 100b + 10c + a$$
$$\overline {cab} = 100c + 10a + b$$
When you add them:
$$\overline {abc} + \overline {bca} + \overline {cab} = 111a + 111b + 111c$$
$$\overline {abc} + \overline {bca} + \overline {cab} = 111(a + b + c)$$
Since $111$ is divisible by $37$, the whole sum is divisible by $37$, but how can i prove that $\overline {abc}$, $\overline {bca}$, $\overline {cab}$ separately are divisible by $37$?
Any tips or hints appreciated.
Best Answer
$$\overline{bca}=10\cdot\overline{abc}-1000a+a = 10\cdot\overline{abc}-27\cdot 37\cdot a$$