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$\overline {abc}$ 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.

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    $\begingroup$ Hint: $111, 222, 333, 444, 555, 666, 777, 888, 999$ $\endgroup$
    – Dilip Sarwate
    Nov 13, 2013 at 12:10
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    $\begingroup$ Consider 4(abc)+7(bca)$. $\endgroup$
    – Gerry Myerson
    Nov 13, 2013 at 12:15

1 Answer 1

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$$\overline{bca}=10\cdot\overline{abc}-1000a+a = 10\cdot\overline{abc}-27\cdot 37\cdot a$$

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