X*X=C can only be 2*2=4 or 3*3=9
Case 1: X=2, C=4 makes the sum
which means L+L=4 or L+L+1=4
the first requires L=2 which is taken and the second is impossible.
Case 2: X=3, C=9 makes the sum
which means L+L=9 or L+L+1=9
the first is impossible but the second requires L=4 (as well as a carry) so we now have
because there is a carry, I+V=12 or 13 depending if there is a carry from the ones place.
Consider I+V=12. The only digits left that satisfy are 5 and 7, neither combination satisfies 3+I=10+V in the ones place.
Consider I+V=13. There are two cases here: 5 and 8, and also 7 and 6.
The combination that fits the ones place is 3+5=8.
So the Arabic numeral solution is
453+485=938 ; 3*3=9
Posted by Jer
on 2011-03-24 11:09:02