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Question

Take a 3 digit number 'abc'. If we add abc, cab and bca together then the resulting sum is divisible by:


A

11

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B

111

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C

37

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D

111 and 37

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Solution

The correct options are
B

111


C

37


The numbers can be written as:

abc = 100a + 10b + c

cab = 100c + 10a + b

bca = 100b + 10c + a

Adding all we get

Sum = (100a + 10a + a) + (100b + 10b + b) + (100c + 10c + c)

Sum = 111a + 111b + 111c

Sum = 111 (a + b + c)

=37×3(a+b+c)

sum111 = (a + b + c)

sum37= 3 (a + b + c)

The the number will be divisible by 111 and 37.


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