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| English, French, and German (Posted on 2010-09-20) |
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Assume that SQUARE, CARRE, and QUADRAT are alphametic squares, that is, that their component letters (accents being ignored) can be replaced by a mutually consistent, non-duplicating set of base-10 digits, with no leading zeroes. Prove that there is in fact no set of numbers by which these letters can be so replaced.
Alternatively, provide a solution to the alphametic.
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Submitted by broll
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Solution:
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Surprisingly, even though only 5 out of 9 different letters are repeated (with only A and R occurring in all 3 words) and C,S,D, and T are ‘wild’, there is no consistent alphametic that can be used to replace the remaining letters.
The problem is soluble without a brute-force search. There are only 16 possible candidates for CARRE between 10000 and 99999; if these rules are then applied to those candidates:
a. RULE 1: If the penultimate known digit is odd, the last digit is 6;
b. RULE 2: If the last digit of SQUARE is 5, then A is 0,2, or 6;
c. RULE 3: If the last digit of SQUARE is 1, then the last 3 digits are e01,e41, e81 if the number of hundreds is even, or o21,o61 if the number of hundreds is odd - corresponding rules apply if the last digit is 9;
then 6 possibilities (excluding repeated digits, etc.) are left for CARRE/SQUARE.
Possible values for T can then be looked up from a table of squares, mod 1000, and the formula x^2-ARE = 1000k, k<1000 used to narrow values for Q and U in QUADRAT, giving 16 'hold-outs' that can easily be checked manually. |
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