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 Near Magic Square Alphametics III (Posted on 2010-03-02)
Substitute each of the capital letters in this 4x4 grid by a different digit from 0 to 9, such that the sum of each of the four columns as well as the sum of each of the four rows is NL.

S   C   A  N
P   O   L   E
A   R   E   A
R   E   S   T

Note: NL represents the number 10*N + L.

 No Solution Yet Submitted by K Sengupta No Rating

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 A different analytic approach Comment 3 of 3 |

As given, L must be 0.  At best, NL as a sum of four digits could only be 10, 20 or 30.   By deleting L (0) and the shared A and E values, A+R+E+A and A+L+E+S can be reduced to S = A+R.  From this, R+E+S+T = R+E+A+R+T = N+E+A+T, which in similar fashion reduces to 2R = N (being 1, 2 or 3).  R can now only be 1 with N = 2, so NL must = 20.<o:p></o:p>

<o:p> </o:p>

The sum of the four row or column totals is now 80.  By subtracting the known N, L and R values, we're left with 3A+C+3E+O+P+2S+T = 76 for the remaining 12 letters.  From REST, E+S+T = 19, and from NEAT, E+A+T = 18.  Thus, E+S+T = E+A+T+1 which reduces to S=A+1.  Also, from <st1:stockticker>AREA</st1:stockticker>, 2A+E must = 19, so E is definitely odd (3, 5, 7 or 9 only).  The only possibilities now for 2A+E = 19 are thus 2(8)+3, 2(7)+5, 2(6)+7 or 2(5)+9.  Subtracting the possible 3A/3E values from 76 yields the following remainders (for C+O+P+2S+T):<o:p></o:p>

<o:p> </o:p>

76-3(8)-3(3) = 43<o:p></o:p>

76-3(7)-3(5) = 40<o:p></o:p>

76-3(6)-3(7) = 37, or<o:p></o:p>

76-3(5)-3(9) = 34<o:p></o:p>

<o:p> </o:p>

For these balances, the following possibilities remain, each of which includes a value for 2(S):<o:p></o:p>

If 43 (with 4, 5, 6, 7 and 9 remaining), the largest possible sum is only 40 (using S=9), which fails.<o:p></o:p>

If 40 (3, 4, 6, 8 and 9 remaining), the largest sum possible, again with S=9, is only 39, which also fails. <o:p></o:p>

If 37 (3, 4, 5, 8 and 9 remaining), only 3+4+5+2(8)+9 = 37 works.<o:p></o:p>

If 34 (3, 4, 6, 7 and 8 remaining), only 3+4+2(6)+7+8 = 34 works.<o:p></o:p>

<o:p> </o:p>

A is now either 6 or 5, and E is either 7 or 9 respectively.  If the former, A+L+E+S would be 6+0+7+7 to = 20, which fails since S can't also be 7.  So A must be 5 and E must be 9, A+L+E+S would be 6+0+9+5 = 20, with S = 5.  With values for N, L, R, A, E and now S determined, the remaining grid values for C(7), O(3), P(8) and T(9) follow quickly to yield:        <o:p></o:p>

<o:p></o:p>

6 7 5 2<o:p></o:p>

8 3 0 9<o:p></o:p>

5 1 9 5<o:p></o:p>

1 9 6 4
 Posted by rod hines on 2010-03-03 14:17:24

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