Listener Crossword 4503 Property Management

[emcee]

I've only just started to look at this. It immediately occurs to me that those fourth powers are also square numbers so surely they will satisfy both properties. But this isn't allowed according to the rubric!

Needs further thought on my part I guess.

[TheBear69]

The instructions do not say that each entry satisfies exactly one property, only that it 'associated' with exactly one property. So if 64 is an entry you can associate it with either a or b (maybe others). You have to end up with 40 'associations' so that the enumerations agree with those specified in the properties list.

I think this is the hardest math listener I've ever encountered....

[emcee]

Thanks for the clarification TheBear69.

I agree with you, it's not going to an easy one.

[crosswhit99]

Presumably if we associate the grid entries at 1a and 16d with property (e) (there are no other candidates for this property) then entries at 4d, 9d, 25a and 36d must be the four required squares ? Is there a reason the preamble states 'exactly one' rather than 'at least one' ?

[crosswhit99]

Sorry, I meant property (c)

[British_Olympics]

Given that there's a unique solution, it's obviously intended to speed up the solving process, in that grid numbers can be eliminated from consideration once they are allocated with certainty to a property eg 28d with (n) rather than (g). Not sure why it's "exactly" and not "at least" though.

[BMACMAC]

I'm confused by the "helpful" information concerning numbers that are the sum of two squares. According to that info, 36 = 2^2 * 3^2 should fit the property since both factors occur twice, but I can't find the two squares that sum to 36, since 6^2 + 0^2 isn't allowed. Anyone help with this?

[TheBear69]

The actual theorem does not exclude 0 as one of the two squares: https://en.wikipedia.org/wiki/Sum_of_two_squares_theorem. As you show, the only way to get 36 is 0^2 + 6^2.

Further, the hint is not enough since it does not help you eliminate numbers like 8 (4+4) unless it were true that the ONLY way to express n=2a^2 as the sum of squares is a^2 + a^2. But 50 = 1 + 49, so that is false. So the hint is useful for eliminate some numbers but not all; i.e. a^2 and 2a^2 have to be checked some other way.

[crosswhit99]

36 is the sum of 6^2 and 0^2, so presumably 'non-zero' should have been omitted in the preamble.

[Scorpius]

That statement also allows 16 since the factors of 16 do not involve a prime that is one under a multiple of 4. But the only squares to make 16 are 4^2 and 0^2.

[Cruncher]

Seem to have run aground. Have about half the grid done but seem to have an ambiguity on the bottom LH corner where either of two entries for 27 would satisfy requirements for the checked across entries. Pants

[Cruncher]

Silly me. Ignore. As you were

[kdgs2]

This was right up my street, hence my willingness to keep at it until the wee hours, although I'm sure it's quite forbidding if you're not particularly mathematical. My solve went in waves, with flurries of progress separated by lots of pencil-chewing.

[Scorpius]

It was a lengthy slog but I'm not sure it was the hardest ever. The maths involved wasn't complicated (unlike some based on large non-decimal bases, one of which I abandoned) and at least I could get some entries in the grid fairly quickly.

[Philoctetes]

For once, in a numerical, it was really possible to take a wrong turn - there were a couple of possible combinations that only revealed themselves to be red herrings when the last two answers refused to fit - rather than make a pig's ear of the sums. Whenever a new sort of "Number" turns up (Mersenne?) I add a list to my spreadsheet, which now runs half way to Cairo. A good work out, this.

[BMACMAC]

Well, I got there in the end, though not without a few false steps. As has been said, the maths weren't all that demanding, more a case of careful application of logic, and plenty of tenacity together with a plethora of table searching. I'm not sure quite how I'd have managed with just a calculator, though!

[perseverer]

There is now a correction on the Listener site - confirming the last sentence of the preamble should not include non-zero.

In retrospect, this puzzle was probably less difficult than I found it!
Radix's base 24 puzzle, number 4256, would get my vote for the most difficult numerical of the last 10 or so years.

[Oyler]

Zag's Codebreaker L4295 in 2014 gets my vote for the toughest numerical in The Times era.

[icynorth]

I had to start from scratch three or four times, but it’s just worked out for me. Yes, a tough puzzle. I didn’t use the hint relating perfect numbers, triangular numbers and Mersenne primes.

Thanks, Smudge!