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:
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.