When a 3-digit number 'abc' is added as abc + bca + cab, their sum is always divisible by .

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Solution

The correct option is A 37
Here,
Let's say a 3 digit number is 'abc',
Now,
Adding 'abc' + 'bca' + 'cab'
= 100a + 10b + c + 100b + 10c + a + 100c + 10a + b
= 111 ( a + b + c)
= $37 \times 3$ (a +b + c)
$∴$ It is always divisible by 37

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