If $a, b,$ and $c$ are three digits of a three-digit number, prove that $a b c + c a b + b c a$ is a multiple of

37

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16

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20

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19

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Solution

The correct option is A $37$
We have $a b c + c a b + b c a$
$a b c = 100 a + 10 b + c$
$c a b = 100 c + 10 a + b$
$b c a = 100 b + 10 c + a$
Adding $a b c + c a b + b c a = 111 a + 111 b + 111 c$
$= 111 (a + b + c)$
$= 37 \times 3 (a + b + c)$ which is a multiple of $37$ .

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