How many positive integers less than 1000 are 6 times the sum of their

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 zero
IF, 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