Listen With Others

Consternation in the Numpty camp when we read the ‘clues’ in IOA’s Prime Cuts. I couldn’t even begin to hunt for any alcohol, though I imagine there had to be a glass of high quality red to go with his prime cuts. Where do we start? ‘Not a prime’, ‘A prime’, ‘A cube’, ‘A square’, ‘A palindrome’? There are rather a lot of all of those.

The other Numpty announced that 16 and 17 could only be 1331 and 11 so we already knew that the 25 primes in the grid would add to 2662. We realized that for 25 primes to fill the 49 cells of the grid, we were going to need all four single-digit primes, 18 2-digit primes (that’s nigh on all of them) and three fairly large 3-digit primes, and that each row of our final grid would have to have a one-digit or a 3-digit prime with 2-digit primes making up the quota.

The initial problem was finding those 7-digit solutions and we laboriously worked our way to 7d, 1ac and 6d, checking that potential solutions would split into convenient primes, and keeping a record of the ones we had used. 232 at 18ac was going to be the sum of the 49 digits (they had to average around 5 and produce a prime ending in 3 + prime 2.) and the three large ones (569, 659 and 587) soon emerged, producing the central palindrome.

But 19ac was the real stroke of genius. It had to be ???1??9 and be the cube of a prime + a higher power of a higher prime. How about 8 + the 6th power of 11? That gives 1771569, two 2-digit primes we haven’t used yet and a large 3-digit prime

Soon we were left with 2d (either 9?1?737 or 9?9?737 and it had to be divisible by each of its digits). I struggled with that one until it occurred to me that I could work out which remaining primes had to complete my grid and sure enough the 172 remaining of those 2662 split conveniently into 3, 19, 61 and 89. Eureka! I checked my digit count, and checked that 1ac was a multiple of 61 (12 across).

No, this wasn’t a puzzle for those who love a mathematical struggle but how clever of IOA to include almost all of the small primes and produce the two totals, to confirm a correct grid, as well as that multiple of 12 across. We’ve been trying to work out who 10ac might be, but have come up with no likely answer. Could he be just a verbal sheep lurking in numerical wolf’s clothing? Thanks to him anyway.