What is $\overline{ABCDE}$ ?

We get a lot of information just considering the last two columns. Let $z$ be the carry-over from the last column and $y$ the carry-over from the second-to-last column, so that we have $7B=10z+A$ , and $6D+z=10y+A$ .

From the second equation, we see that $z-A$ must be even - that is, the two digits of $7B$ must have the same parity (note that if $B<2$ we have $z=0$ ). The only such multiples of $7$ for $B<10$ are $7\times 0=00$ , $7\times 4=28$ , $7\times 5=35$ and \ $7\times 6=42$ .

Clearly, we can't have $B=0$ , because then we'd need $A=0$ , which isn't allowed. Similarly, if $B=5$ we get $A=5$ too, not allowed. Let's look at the other cases:

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Case $B=6$ : we get $A=2$ and $x=4$ . So $6D+4=10y+2$ , and $D$ is either $3$ or $8$ , giving $y=2$ or $y=5$ respectively.

The next equation we have is $5E+y=10x+A$ . Plugging in the two cases, we find either $5E+2=10x+2$ or $5E+2=10x+5$ . The second of these is clearly impossible; so the only subcase to consider is

$B=6,A=2,D=3$ and $E$ even. The only remaining even numbers are $0$ , $4$ and $8$ , giving $x=0$ , $x=2$ and $x=4$ respectively.

Checking the next equation, $4D+x=10w+A$ , we already know $D=3$ and $A=2$ ; substituting in, $12+x=10w+2$ . So $x=w=0$ and $E=0$ .

$3C=10v+2$ ; only solution is $E=4$ , $v=1$ .

$2B+v=10u+A$ ; substituting, $12+1=10u+2$ , which doesn't work! So we can't have $B=6$ .

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Case $B=4$ : we get $A=8$ and $x=2$ . So $6D+2=10y+8$ , and $D$ is either $1$ or $6$ , giving $y=0$ or $y=3$ .

$5E+y=10x+8$ ; so the only possibility is $D=6$ , $y=3$ , and $5E+3=10x+8$ so that $E$ must be odd. We haven't used any odd digits yet, so the only information this gives us is $0 \leq x \leq 4$ .

Next equation: $4D+x=10w+A$ ; substituting in, $24+x=10w+8$ ; so we must have $x=4$ , $w=2$ and $E=9$ .

$3C+w=10v+A$ ; substituting, $3C+2=10v+8$ ; so the number $3C$ ends with a $6$ - ie $C=2$ .

Pulling all this together, we find the only possible solution is $\boxed{\overline{ABCDE}=84269}$ , and checking in the full equation we see this does indeed work.