abc, bca and cab are the three numbers formed using three digits a, b and c. Addition of all the three numbers gives a new number. The new number is divisible by _____.

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Solution

The correct option is C 37
$Here, a b c = 100 a + 10 b + c$
$b c a = 100 b + 10 c + a$
$c a b = 100 c + 10 a + b$

$So, a b c + b c a + c a b = (100 a + 10 b + c) + (100 b + 10 c + a) + (100 c + 10 a + b) = 100 a + 10 a + a + 100 b + 10 b + b + 100 c + 10 c + c = 111 a + 111 b + 111 c = 111 (a + b + c) = 37 \times 3 (a + b + c)$

Therefore, the number is divisible by 37.

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