How to solve a algebraic equation?

Question

My maths teacher gave me this equation and I really don't know how to solve this: $$\overline{abc}+\overline{ab}+\overline{bc}+\overline{ac}+a+b+c=29,$$ where $a$, $b$, $c$ are digits.

I need to find how many three digit numbers are there which satisfy this.

Answer 1

You have three equations in three unknowns. From the least significant digit (take the rightmost digit of each term) you have

$$c + b + c + a + b + c = 9$$

From the next digit you have $$b + a + b + a = 2$$

From the most significant digit you have $$a = 0$$ noting that $29 = 029$.

Can you take it from here?

Answer 2

Well write it as $100a+10b+c+10a+b+10b+c+10a+c+a+b+c=29=121a+22b+4c$ , so you get $a=0$ but also no integers for $b$ and $c$ satisfy it. so no solution.