## ‘Can’t You Do Division’ by Oyler

Posted by Encota on 10 Mar 2017

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*, 2*x*, 3*x*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

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