A three digit number is 12 times the sum of its digits. What can the number be?  

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We are given that 12 times the sum of the digits of a three digit number is equal to the number. Let the number be abc where the digits of the number are a, b and c.

Now (a + b + c)*12 = 100*a + 10*b + c

=> 12a + 12b + 12c = 100a + 10b + c

=> 88a - 2b – 11c = 0

Now a can only be 1 as 2b + 11c can take on a maximum value of 9*2 + 9*11 = 117 and 88*2 = 176.

So, we now have to find values of b and c such that 2b + 11c = 88

11c can take the values 88, 77, 66…

For 11c = 88, we can take b = 0. For 11c = 77, there are no options for b. The same goes for 66 and other decreasing values of c.

The required number is therefore 108.

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