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Twice true (Posted on 2011-03-24) Difficulty: 3 of 5
The above equations are true both as Roman expressions and as alphametics, where each letter denotes Arabic numeral. What is the unique solution?

See The Solution Submitted by Ady TZIDON    
Rating: 4.5000 (2 votes)

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Solution 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
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
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