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MatydaLover | 20:39 Sun 27th Jan 2008 | Quizzes & Puzzles
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Another tough one this week. Nevertheless, anyone else finding it elementary so far?

p. s. Sorry to miss last week's discussion - my login was repeatedly rejected until Tuesday. Seems ok now, though.
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As Homer (Simpson) would say, DOH! Thanks very much. Will persevere.
Totally agree with Slaney. It really is worth persevering with. Also agree with the previous comments about the ingenuity ( and sheer sadism! ) of the compilers. If any of them read this forum I'd like to know approximately how long on average it takes to compile such a puzzle.
Have done it! Thanks for the hints, have a few questions regarding my answers but assume they're right as they allow completion. First completion for about 3 months!! (been too busy). Spotted the square to be replaced by "2" almost directly after posting my last comment and then accelerated rapidly to the solution.
Thanks again
The most straightforward way to solve the square is to start with the upper left blank cell - this must be the difference between the sum of the three numbers in cells 14, 29 and 36 and the sum of the two numbers in cells 26 and 31. Everything falls into place thereafter.
More like the tortoise than the hare this week but have now finished and feel it well worth the effort. I was fine with the rotational symmetry and most of the clues but really had a job visualising the square, so a bit of trial and error on getting the appropriate 'elementary' collections together
Question Author
That's good news! I think it's great that so many people have persevered and succeeded- the satisfaction from completing this one is almost boundless.

Late Sunday night I almost gave in ... "aw, forget it, It's only one". Then I thought about all the people here who would NOT give in - so thanks to all for keeping me going.
Firstly I'd like to thank everyone who has helped me so far - I have enjoyed it all. I thought it was going to be pretty easy.

However I now have trouble with the magic square:
all rows and columns add up to 127; but my two diagonals both add up to 126.
Diagonal a1-c4 is 61-22-30-13 &
diagonal c1-a4 is 50-33-41-2

Where have I gone wrong?
Question Author
I'm at work right now and I don't have my copy with me. If no-one else does so before, I'll post this evening. But meanwhile from memory ...

Use the top row and the diagonal from top right to bottom left. You get nnn + A = nnn + A + B; the A's cancel, so you get B. (B is the bottom left square.)

I think the numbers were ... 114 + A = 52 + A + B, therefore B = ... ... and the rest should follow.

The magic constant is close to 127.

As MatydaLover says, the magic constant is very close to 127 so try one or two on either side of it and you should get it.
offramp, each of the 'unknown values' you have used in a diagonal needs increasing by 1, so your total increases by 2. For example, the cell you have filled with 61 must be the difference between the sum of cells 11, 14 and 19 minus the sum of cells 26 and 29, that is 114 minus 52 = 62. This then sets the total at 62 plus the sum of cells 36,39 and 44.
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If you're still in any doubt, then your grid should look like this:

2-----54-----58-----A
*------*------30----18
26----22-----*------*
B-----10-----6------50

As in previous post: 2 + 54 + 58 + A = B + 22 + 30 + A
Subtract A from each side and you are left with ...
2 + 54 + 58 = B + 22 + 30

All the best!


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