 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Twice true (Posted on 2011-03-24) LIX+LVI=CXV ; X*X=C
The above equations are true both as Roman expressions and as alphametics, where each letter denotes Arabic numeral. What is the unique solution? Comments: ( Back to comment list | You must be logged in to post comments.) full solution | Comment 1 of 3
X*X=C can only be 2*2=4 or 3*3=9

Case 1: X=2, C=4 makes the sum
LI2+LVI=42V
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
LI3+LVI=93V
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
4I3+4VI=93V
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 Please log in:

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