Imagine a rectangle divided into 3x4 squares, and put a digit in each square.
+++++
 a  b  c  d  A
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 e  f  g  h  B
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 i  j  k  l  C
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D E F G
The number abcd is denoted by A, that is, A = 1000a + 100b + 10c + d, and the same for the other 2 horizontal numbers B and C.
The number aei is denoted by D, that is, D = 100a + 10e + i, and the same for the other 3 vertical numbers E, F and G.
Prove that if any 6 of these numbers (A, B, C, D, E, F, G) are divisible by 7, then the last number must also be divisible by 7.
10 is congruent to 3 mod 7.
100 is congruent to 2 mod 7.
1000 is congruent to 6 mod 7.
We are only concerned with the congruences mod 7, so only the value of the digits mod 7 need be considered. For example, it doesn't matter whether a is 1 or 8 as it's only a difference of 7000, which is divisible by 7, and either way its value to the overall congruence of ABCD is 6. Similarly if a were 2 or 9, it would be worth 5 either way.
In testing the hypothesis for, say, all the rows and the first three columns, a, b and c determine what d must be to satisfy one of the premises. Likewise e, f and g determine h, and similarly for the last row. For the columns, a and e determine i, etc.
In this particular case, l can be determined from either the previously determined d and h, or the previously determined i, j and k. If they don't agree, the hypothesis is disproved.
The following program, for all combinations of a, b, c, e, f and g, determines the necessary d, h, i, j and k. It then sees if the l value derived from d and h is equal to the l value derived from i, j and k. If the l values are not equal, that is reported. Capital letters are used for the variables due to UBASIC syntax; not to be confused with the capitals in the puzzle statement.
10 for A=0 to 6
20 for B=0 to 6
30 for C=0 to 6
40 for E=0 to 6
50 for F=0 to 6
60 for G=0 to 6
100 D=(14(6*A+2*B+3*C)@7)@7
110 H=(14(6*E+2*F+3*G)@7)@7
130 I=(14(2*A+3*E)@7)@7
140 J=(14(2*B+3*F)@7)@7
150 K=(14(2*C+3*G)@7)@7
160 L1=(14(2*D+3*H)@7)@7
170 L2=(14(6*I+2*J+3*K)@7)@7
180 if L1<>L2 then
190 :print A;B;C;D:print E;F;G;H:print I;J;K:print
999 next:next:next:next:next:next
For all 117,649 combinations of a, b, c, e, f and g, no discrepancy is found.
At first glance this would indicate we've only checked one case, or rather two cases: where the right column (G) was the dependent one, or the bottom row (C) was.
However, 7 is a prime number, making it, in particular, relatively prime to 6, 2 and 3, the coefficients of the digits used in adding up the values of the entire 3 or 4digit numbers mod 7. As a result, there is a 1to1 correspondence between, say, the needed a values and the needed d values to make ABCD = 0 mod 7. So if, say, column 7 is a known multiple of 7, each digit there determines, along with the middle two columns, what column a is. And we have already determined that combinations of the first 3 columns determine the rest.
In fact, this property was already used in our indifference as to whether the "remaining" number was the bottom row or the righthand column. In either case it did not matter what column/row (respectively) was used as the test for compatibility.

Posted by Charlie
on 20081104 13:19:14 