Listen With Others

So, before attempting Oyler’s excellent puzzle, did you already know that, if a number has a prime number of factors then it has to  be of the form (prime) raised to the power n?

“Not a lot of people know that…”

As an aside, I’ll use ^ to mean raised to the power of, rather than powering up a maths typesetter!  So 2^6 = 2x2x2x2x2x2 = 64.

I loved the idea that, if the clue was a prime number – say 31 – then the only possible answer is (prime) to the power 30.  So if the answer was 10 digits it’d have to be 2^30 = 1073741824.  So much info hiding in such benign-looking clues!

Armed with this, then it’s easy to make quick work of:

K     7  (2)

S     5  (2)

K must be a 6th power of something and only 2^6 = 64 fits; S must be a 4th power so could be 2^4 or 3^4, i.e. 16 or 81.  As t is prime (since it only has two factors, itself and 1) then it must be odd and so S is 16.  We’re off!

Moving on to clue p.  It has 35 factors, so it could be prime^34 but even 2^34 is way too big.  The only other way to factorise it (apart from 35×1) is 5×7, so it must be of the form (primeX)^4 x (primeY)^6.  And the only way to make that small enough is 3^4 x 2^6 = 5184.

As usual with numericals I filled it out using my high-tech rainbow-inspired range of pencils – i.e. the red entries were those I put in first, then green and finally blue – to make it easier to re-work if I made a mistake.  I personally took the solving route that focused on the clues with a given number of factors then let the perimeter of perfect squares drop into place, though of course it would have been easy (computer-aided) to create the full list of ‘123456789’ jumbles that are perfect squares and head along that path.

I did waste 90 minutes when I thought I had an error at entry f.  It turned out I had been looking for a number less than 691 to fit in there, when I should have been looking for a number less than or equal to 691.  I started again from scratch, only to find that the missing answer was 691.  Grrr!

Knowing some of Oyler’s work, I wouldn’t be surprised to find one or two other things hiding in the puzzle:

• I see Rhombus’ original puzzle number of 1916 features in the corners of the puzzle, e.g. clockwise from top right.
• The centre column not only contains three 3-digit numbers but 1 x 192, 2 x 192 and 3 x 192, the only(?) x, 2x, 3x triple where this is the case!  [And entries at B, K and S are all factors of these]
• I bet there’s at least one more item hidden in there somewhere.  I looked for 4438, the puzzle number, but without success.  And, forgive me, I looked for it in various bases too!  I was then tempted to find the HARE but my calculator ran out of digits multiplying 38 x 672935481 x 234975 x 8132 so I had no way of working out 48862541494458642600.

And, finally, it may interest the Listener historians amongst you that the answers at m and p are identical in both the 1967 puzzle and this 2017 tribute.

Fabulous puzzle – thanks Oyler!

cheers,

Tim / Encota