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rad

19th May 2018, 00:03
Another preamble error, I fear. In the last sentence the word non-zero should be omitted. For example, 9 is not the sum of two non-zero squares.
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verlaine

19th May 2018, 08:23
Interesting observation Rad! Can we at least assume that any integer that can possibly be an answer in this puzzle (i.e. not single-digiters) will follow this rule? As a non-mathematician I couldn't possibly tell you the answer to that, but maybe someone will know.
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verlaine

19th May 2018, 08:26
Maybe not I guess? 16? I'm so confused.
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gitto

19th May 2018, 08:34
I am certainly confused, as I do not see how there can be a solution for property e. If I've understood the preamble properly, which I now think is extremely doubtful, e is either 2^2,3^3,4^4 etc. None of these gives a solution of the right number of digits to fit the entry's grid number. It seems pointless to continue as I clearly do not understand what to do!!
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rad

19th May 2018, 10:39
Verlaine

Property f cannot be used for entries at 9 or 36. Entries with property f cannot be squares of primes such as 7, 11, ...
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rad

19th May 2018, 10:43
Gitto

The entry with property e could go in at 1, 4 or 27. It could be 4 to the 4th, 5 to the 5th, etc.
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crosswhit99

19th May 2018, 10:49
… except there seems to be another problem with the preamble. If the entries at 1a and 16d are fourth powers (there are two and no other grid numbers are fourth powers), then they must also be squares, so these two entries are not associated with exactly one of the properties as stated in the preamble's first sentence.
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dryden

19th May 2018, 11:07
9 is ruled out by the rest of sentence because the factor 3 (one less than a multiple of 4) occurs an odd number of times in the factorisation.

The opening sentence of the preamble doesn't meant an entry or grid number coulddn't be associated with more than one entry in theory. There are 6 possible grid numbers satisfying a; there are two satisfying c. I'm assuming that the two that satisfy c are ruled out of consideration for a.

I may be wrong, of course. I'm pretty confused myself. I do have 11 tentative entries that fit together, though I have only three of which I'm pretty certain.


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crosswhit99

19th May 2018, 11:34
It looks like the intended reasoning is that there are only two grid entries that can satisfy property (c), and after eliminating their grid numbers from the possibilities for properties (a) and (e), for example, the grid numbers for these are also identified and 27d can be entered with certainty.
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crosswhit99

19th May 2018, 11:36
Sorry Dryden, I didn't see you reply and it looks like we concur !
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