the sum of the digits of a three digit number is six. the ten digit is one less than the units digit and the number is twelve more than one hundred times the hundred digit

Find the three digits
x
y=x-1
z=7-2x

x+x-1+z=6
2x-1+z=6
z=7-2x

x-1=100(7-2x)+12
what is wrong with my reasoning?

It says "the number is twelve more than one hundred times the hundred digit"

The number is 100z+10y+x. So, you want

100z+10y+x = 100z+12
Now plug in your values and solve for x.

Your reasoning is almost correct, but you made a mistake in the equation:

x - 1 = 100(7 - 2x) + 12

Let's break it down step by step to find the error:

1. Start by considering the equation for the tens digit and the units digit:
y = x - 1

2. Next, let's consider the equation for the hundreds digit:
z + 12 = 100x

3. Finally, let's consider the equation for the sum of all three digits:
x + y + z = 6

Now, substitute the values of y and z into the equation for the sum:

x + (x - 1) + (100x - 12) = 6

Combine like terms:

102x - 13 = 6

Add 13 to both sides:

102x = 19

Divide both sides by 102:

x = 19/102

So, according to your problem statement, the value of x is a rational number, which contradicts the fact that you are looking for a three-digit number.

Therefore, there appears to be a mistake in the problem statement or the equations provided. Please double-check the information given and make sure there are no errors or missing details.