Question 882838
Looking at the ones digits:
{{{0<=B<=9}}}
{{{9+3+0<=9+3+B<=9+3+9}}}
{{{12<=9+3+B<=21}}}
Since the ones digit of {{{9+3+B}}} is {{{3}}} ,
{{{9+3+B=13}}}-->{{{12+B=13}}}-->{{{B=13-12}}}-->{{{highlight(B=1)}}}
When adding the tens digits, we have the {{{1}}} ten from {{{9+3+B=13}}} carried over.
{{{1+C+7+6=10A+A}}}
{{{C+14=(10+1)A}}}
{{{C+14=11A}}}
So {{{C+14}}} is the multiple of {{{11}}} {{{11A}}} .
Since {{{0<=C<=9}}} ,
{{{0+14<=C+14<=9+14}}}
{{{14<=C+14<=23}}}
The only {{{C+14=11A}}} multiple of {{{11}}} between {{{14}}} and {{{23}}} is
{{{11*2=22}}} .
So {{{11A=11*2}}}--->{{{highlight(A=2)}}} and
{{{C+14=22}}}--->{{{C=22-14}}}--->{{{highlight(C=8)}}} .
Then, {{{A-B+C=2-1+8}}}--->{{{highlight(A-B+C=9)}}} .