How to find $\overline{abc}_{7}-\overline{cba}_{7}=\overline{xy4}_{7}$?

Question

The problem is as follows:

Let:

$\overline{abc}_{7}-\overline{cba}_{7}=\overline{xy4}_{7}$

Assuming $a=b-c$

Find: $cba_{7}$ in terms of decimal system.

How exactly should I do this?. Can someone help me here?.

The thing is I don't know how to deal with carry numerals.

So far the only thing which I spotted is that $c-a=4$

For this the only way to get this is assuming $c=6$ and $a=2$ thus it will not carry. Isn't it?.

But this is somewhat contradictory by looking on the third digit because it states $a-c=2-6=-4$

Which it cannot happen. Thus what am I doing wrong?. Can someone help me here?. My only requirement is a method which doesn't really uses much sophisticated math such as modular arithmetics or too many algebraic manipulations. Does it exist a way to do this more intuitively?

Answer 1

BIG HINT

It is important to see that $c-a$ is not $4$. From the leading digit one has that $a$ is at least as big as $c$. So the only possibility is that $7+c-a=4$ i.e. there has been a carry. We therefore obtain $a=c+3$.

You should now be able to see that $\overline{abc}$ must be treated as $$\overline{(a-1)(b+6)(c+7)}$$.

The rest should be easy but I will add that there seems to be a convention that no leading digits are zero.

Answer 2

$$a,b,c,x,y \in \{0,1,2,3,4,5,6\}$$

$$abc-cba7 = xy4$$

$$abc = cba + xy4$$

So we want to end up with this addition result (in base 7).

\begin{array}{c} & c & b & a \\ + & x & y & 4 \\ \hline & a & b & c \end{array}

Let's start with the right column

\begin{array}{c} & c & b & \color{red}a \\ + & x & y & \color{red}4 \\ \hline & a & b & \color{red}c \end{array}

Either {$a+4=c$} or {$a+4 = 7+c$, which implies $c=a-3$}.

From the last column, we conclude that $c < a$. So $c = a-3$ and, because $a+4 = 7+c$, we must carry a $1$.

\begin{array}{c} & c & \color{red}{\overset{\displaystyle 1}{b}} & a \\ + & x & \color{red}y & 4 \\ \hline & a & \color{red}b & c \end{array}

So, either $(b+1)+y = b$ (which implies $y=-1$) or $(b+1)+y = 7+b$ (which implies $y=6$.

We conclude that $y=6$ and we must carry a $1$.

\begin{array}{c} & \color{red}{\overset{\displaystyle 1}{c}} & \overset{\displaystyle 1}{b} & a \\ + & \color{red}x & 6 & 4 \\ \hline & \color{red}a & b & c \end{array}

So it must be true that $(c+1)+x = a$. Substituting $c=a-3$, we conclude $x=2$.

\begin{array}{c} & \overset{\displaystyle 1}{c} & \overset{\displaystyle 1}{b} & a \\ + & 3 & 6 & 4 \\ \hline & a & b & c \end{array}

So, if you want unique answers

$a=3 \qquad b =5 \qquad c=0 \qquad x = 2 \qquad y=6$

\begin{array}{c} & 0 & 5 & 3 \\ + & 2 & 6 & 4 \\ \hline & 3 & 5 & 0 \end{array}

$$cba_7 = 053_7 = 38$$

$$xy4_7 = 264_7 = 144$$

$$abc_7 = 350_7 = 182$$