# 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.**

Let the digits of the 3 digit number be x,y and z.

Then the value of the number is 100x+10y+z which should be 12 times the sum of the digits implies:

100x+10y+z= 12(x+y+z).

Therefore (100-12)x+10x+z =12y+12z.

=> 88x= 12y-10y +12z-z.

=> 88x = 2y+11z.

=> If x = 1, y = 0, then z = 8 satisfies the relation 88x=2y+11z.

=> So the required number is 108 .

Therefore the required number is 108.