## Listener No. 4373: *Semirp* by Kea

Posted by Dave Hennings on 11 December 2015

A second mathematical puzzle from the normally word-oriented Kea, his first being *4164* back in November 2011. If memory serves me, that was fairly straightforward. I wondered if he had used the intervening four years to come up with something a little more tricky. For a start, it seemed there were only 15 clues. Closer inspection showed that there were actually only 7, the others just telling us to look at another one. (In fact, as I would suss later, these *were* clues since they ensured the referenced one was distinct.)

Every entry was the reverse of a prime, so I put 1, 3, 7, 9 in the cells that were the first digit of an entry, being the options for those cells. The 2-digit entries B, F and M now had those options in both digits so were 13, 17, 31, 37, 71, 73, 79, 91, 93 or 97. 93 was not the reverse of a prime and 91 was the only non-prime which enabled distinct values of B and N in clue I *(B + B) / N* where N ≠ 2.

Thus 2B was 182, which factored to 1.2.7.13 leading to N being 13 or 14 (26 not reverse of a prime and 91 not distinct). If N = 13, then I is ·3, ie 13, 33, 73 or 93, only 73 being distinct, with no consecutive digits the same and reverse of a prime; Y = 59. If N = 14, then I = ·4, only 14, 34 and 74 being valid; Y = 34, 74.

Valid values of D with middle digit 9 and non-prime are 391, 393, 791 and 793. Only 793 is divisible by one of the values for Y, ie 61, giving T = 13, N = 14 and I = 74.

O *NUN* came next, with only 392 fitting the rules and U = 2. From A *D + O / W* we get A = 797 and W = 98.

Using H *SC / F*, SC / 17 = 713. SC = 17.23.31 so S = 23 or 31 and C = 391 or 527, ie 391.

Using adjacent digits, E = 904, 924, 934, 954, 964 or 984. Only 904 and 934 are reverse of a prime. Using adjacent digits, K begins with 3, so M = 73 and J = 34, 35, 37 or 38.

Inspecting the last digits of *J + J – G + L*, we get 35 + 35 – 1·9 + 152 = 73 and G = 149.

Finally, E *(K – L + U – T)Z* with E = 904 (2.2.2.113) or 934 (2.467). 467 is too large with K = 3··, as are 452 and 904, and 113 too small. Thus K =369, giving (369 – 152 + 2 – 13).4 = 904.

Not too tricky a puzzle in the end from Kea, but an entertaining couple of hours in the company of primes, thanks.

The final table of all values is:

A | B | C | D | E | F | G | H | I | J | K | L | M | N | O | S | T | U | W | Y | Z | |

797 | 91 | 391 | 793 | 904 | 17 | 149 | 713 | 74 | 35 | 389 | 152 | 73 | 14 | 392 | 31 | 13 | 2 | 98 | 61 | 4 |

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