Semirp by Kea

Puzzle solution process

This is one possible solution path. As they are the reverse of primes, all entries start with one of {1, 3, 7, 9} and A, B, D, F, H and M consist of those digits only. The two-digit entries in that list are B, F and M, for which the possible values are {13, 17, 31, 37, 71, 73, 79, 91, 97} (not 11, because it would have the same digit in adjacent cells); note that these are all prime numbers except for 91 = 7×13.

From O = NUN, N2 has no more than 3 digits, so N < 32. From I = (B + B)/N + Y, N is a factor of 2B. If B is prime then the only factors of 2B are {1, 2, B, 2B}; N can’t be 1, 2 or B, so N would have to be 2B, which means B < 16, ie, B = 13 and N = 26. But 26 isn’t the reverse of a prime, so B isn’t prime. Therefore B = 91. That gives 9 for the middle digit of D and restricts its first digit to 3 or 7 (having neighbours of 9 and 1).

The factors of 2B = 2×7×13 are {1, 2, 7, 13, 14, 26, 91, 182}, of which N can only be 13 or 14. If N = 13 then F = 17 (the only other available value starting with 1), which would make I = 73, and then Y = I − 2B/N = 73 − 2×91/13 = 59. D = YT is a multiple of Y, but the only multiples of 59 in the 300s or 700s are {354, 708, 767}, none of which matches D, so N = 14.

F is 13 or 17. If F is 13 then (from the grid) I is 34, which makes Y = 34 − 2×91/14 = 21. The only odd multiples of 21 in the 300s or 700s are {315, 357, 399, 735, 777}, none of which matches D (399 does, but it has two adjacent digits the same), so F = 17, I = 74 and Y = 61. The only odd multiples of 61 in the 300s or 700s are {305, 793}, so the only match is D = 793, giving T = 13.

A = D + O/W tells us A > D, and as it has the same first digit and contains only digits {1, 3, 7, 9}, it must be A = 797, with W = O/4.

H is now 71_ and its last digit is one of {3, 7, 9}, which also starts O. We know O = 196U. The only multiples of 196 in the 300s, 700s or 900s are {392, 784, 980}; 980 isn’t allowed (it ends in zero); 784 would make H = 717, in which the sum of digits is a multiple of 3, so 717 (forwards or backwards) is a multiple of 3, not a prime. Therefore O = 392, U = 2, H = 713 and W = 98.

From H = SC/F we now have SC = FH = 12121, so S and C are both odd, not ending in 5, ie each of them ends in one of {1, 3, 7, 9}. Thus C is in the range [313, 397] (its middle digit can’t be 0 because it’s the last digit of G), so S is between 12121/397 ≈ 30.5 and 12121/313 ≈ 38.7, ie it’s one of {31, 33, 37}. The only one that gives an integer result is S = 31, with C = 391. In the grid, M is 7_ but its second digit has neighbours of 1, 7 and 9, so M = 73.

For M = J + J − G + L, we have M = 73, G ending in 9 and L ending in 2, so 2J = M + G − L ends in 0. As J itself can’t end in 0, it must end in 5, so J = 35, completing L = 152 in the grid. Then G = 2J + L − M, so G = 149.

E = (K − L + U − T)Z now reduces to E = (K − 163)Z. K is 3_9 in the grid; its middle digit can’t be any of {0, 3, 6, 9} because they’d give a digit sum that’s a multiple of 3, so the entry wouldn’t be the reverse of a prime, and it can’t be any of {2, 3, 4, 5, 7, 9} because those digits occur in neighbouring cells; so K is 319 or 389 and E (9_4 in the grid) is a multiple of 156 or 226, respectively. The only multiple of 156 in the 900s is 936, not a match for E, so K = 389. The only multiple of 226 in the 900s is 904, so E = 904, Z = 4 and the puzzle is complete.

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