Bunuel
How many positive integers less than 1000 are 6 times the sum of their digits?
(A) 0
(B) 1
(C) 2
(D) 4
(E) 12
Let x = the hundreds digit of the integer
Let y = the tens digit of the integer
Let z = the units digit of the integer
So, 100x + 10y + z = the VALUE of the integer
And x + y + z = the SUM of the digits
We want the VALUE of the integer to be 6 TIMES the sum of its digits
So, we want: 100x + 10y + z = 6(x + y + z)
Expand: 100x + 10y + z = 6x + 6y + 6z
Subtract 6x from both sides: 94x + 10y + z = 6y + 6z
Subtract 6y from both sides: 94x + 4y + z = 6z
Subtract z from both sides: 94x + 4y = 5z
IMPORTANT: Since x, y and z are DIGITS (from 0 to 9), we can see that
x must equal zeroIF, for example, if x has a non-zero value like x = 1, we get: 94 + 4y = 5z, and there are no DIGIT values of y and z that can satisfy the equation.
So, x must be zero.
We get: 4y = 5z
Since y and z are DIGITS (from 0 to 9), we can see that there is only
ONE solution: y = 5 and z = 4
So, 54 is the ONLY positive integer that satisfies the condition that the number is 6 times the sum of its digits
Answer: B
Cheers,
Brent