Puzzle explanation

The odd digits in the completed grid were highlighted, producing the form of one of the creatures from the 1978 arcade game Space Invaders.

Puzzle solution process

This is one possible solution path. The last digits of T and W form w, which is a two-digit fourth power, ie one of {16, 81}, so X = T + W ends in 7 or 9, and as it’s a palindrome, X is one of {77, 99}. But x is a square and therefore can’t end in 7, so X = 99 and w = 81. The only square that fits x is x = 49, which makes the palindrome U = 44.

s = 9S is now _49 in the grid, and to be a multiple of 9 its digits must add up to a multiple of 9, so it must be s = 549, making S = 61 (prime, as expected). Now v is 1_ and the only square that fits is v = 16. For Q we have 5_ in the grid and the only triangular number in the 50s is Q = 55.

We have T = _8 and W = _1, and they add up to 99, so their first digits must add up to 9. q has increasing digits, so it must match one of {_18, _27, _36, _45}, starting with one of {1, 2, 3, 4}. But N is a prime, so it ends in 1 or 3, and it has the same digit sum as Q, ie 10, which allows only {91, 73}; 91 = 7×13 isn’t prime, so N = 73. Now q is one of {336, 345} with T as one of {38, 48}; T is divisible by its digit sum, but 38 isn’t divisible by 3 + 8 = 11, so T = 48, making q = 345 and W = 51.

From the grid, n is __5 (an odd multiple of 5) and it’s a multiple of x = 49, so it’s one of {245, 735}. But we know A = n − K, where K is an anagram of n, so n > K and therefore n can’t be 245; that leaves n = 735.

K is one of {357, 375, 537, 573}, which when subtracted from n give {378, 360, 198, 162} respectively for A, but we can rule out the second of those because c can’t start with 0. Now c is a three-digit multiple of 49 starting with 2 or 8, ie one of {245, 294, 833, 882}, for which the respective digit products are {40, 72, 72, 128}. But because z is a two-digit multiple of c’s digit product, c’s digit product must be < 50, so c = 245, z = 80, A = 162 and K = 573.

We know d = 2×55×735 + a ends in 7, so a ends in 7, and because its digit product is 0 it must be a = 107, which makes d = 80957. Then we have 59_ for the palindrome D, so D = 595.

M = 2f has two digits, so f < 50; the possible triangular numbers are {10, 15, 21, 28, 36, 45}. C is a palindrome (matching _404_) and can’t start/end with 0, so 10 is ruled out; b has the same last digit as f and can’t be the prime 61, which rules out 21. C is a multiple of f and the four possibilities for C/f are {54045/15 = 3603, 84048/28 ≈ 3001.7, 64046/36 ≈ 1779.1, 54045/45 = 1201}, so C = 54045, b = 65 and f is one of {15, 45}.

For V we have _8, which is the product of its own two digits plus 18 (the digit sum of C). That means V’s digit product ends in 0, which is only satisfied by V = 58 (and 5×8 + 18 = 58, as expected). Now t = 55×16 + 58, ie t = 938. Then we have _9 for R, which is a factor of 58, so R = 29. Now o ends in 2 and because L is prime o must start with one of {1, 3, 7, 9}; it can’t be 92 = 23×4 because it’s the digit product of r, which can’t include a two-digit prime. The three possibilities for B = o×(digit sum of o) + 1 are {12(1 + 2) + 1 = 37, 32(3 + 2) + 1 = 161, 72(7 + 2) + 1 = 649}, but we know its middle digit has to be 1 or 4, so B = 649 (completing e = 645), o = 72 and f = 45, making M = 90.

For L we have _37, so its digit sum is in the range 11 to 19. The only Fibonacci number in that range is 13, so L = 337. For i we have __3 and it’s 735 plus a Fibonacci number, so the Fibonacci number ends in 8 and is ≤ 993 − 735 = 258. The only matching one is 8, so i = 743.

E = 55u − 5×649 is an obvious multiple of 5 and the first digit of k can’t be 0, so E ends in 5, u is even, and k, having a repeated digit, is one of {557, 577}. J is 16 more than a cube, for which the three-digit options are {141, 232, 359, 528, 745}; but its middle digit has to be 5 or 7, so J = 359 and k = 557.

A = 162 and j have the same digit product, ie 12, so j is one of {26, 34, 43, 62}. The even values of u from 60 to 68 give {55, 165, 275, 385, 495} for E, of which only the last three could fit with j, so u is one of {64, 66, 68}. F = u + 1, ie one of {65, 67, 69}, which gives _69 for the palindrome h, so h = 969. P = q + y = 345 + 5_, in the range 395 to 404, but r can’t start with 0, so P is in the range 395 to 399. The possibilities for l are now {553, 753, 953}. Any triangular number Z is of the form Y(Y + 1)/2, where Y and Y + 1 are the consecutive integers either side of √(2Z). If Z is 1553 then Y is 55, but 55×56/2 = 1540, so 1553 isn’t a triangular number; if Z is 1753 then Y is 59, but 59×60/2 = 1770, so 1753 isn’t triangular; that leaves 1953, with Y = 62, and 62×63/2 = 1953, so it is triangular. Therefore l = 953, F = 69, u = 68 and E = 495, making j = 43.

m is a multiple of a = 107 with 3 as the middle digit, for which the only possibility is m = 535, making G = 34 and P = 395. y = P − q, so y = 50.

We know that o = 72 is the digit product of r, which starts with 9, so its other two digits must have a product of 8; since its digits are in descending order, r is one of {942, 981}. As t = 938 is the mean of p and r, p and r are the same distance from 938, in opposite directions; if r is 981 = 938 + 43, then p = 938 − 43 = 895, but we know from the grid that p starts with 9. Therefore r = 942 and p = 934.

The grid is now complete apart from H = 7__, the two-digit * entry, and g = 9__. The only values g could have, being 1 more than another entry, are {935, 939, 943, 954, 970}. From G’s clue we know that g’s digit product is a multiple of 54, so the product of the second and third digits must be a multiple of 6, which is only satisfied by g = 943.

H matches 7__ and the sum of the digits entered so far is 364, so H = 728 + 2×(the sum of the 3 unknown digits), and from v’s clue we know it’s a multiple of 16. The * entry is less than u = 68, so H must end with one of {2, 4, 6}, the possibilities being {736, 752, 784}. (In fact, if we were to look ahead at decoding the grid, the second pair of digits is 28, which can yield a letter from 1 to 26 only if * is less than 30, forcing H to be 752.) If H is 736 then the missing second digit of * would have to be 736/2 − (364 + 3 + 6) = −5, not possible; if H is 784 then the missing digit would have to be 784/2 − (364 + 8 + 4) = 16, too big; therefore H = 752 and the missing digit is 752/2 − (364 + 5 + 2) = 5, making * = 25. This completes the grid.

Decoding the digit pairs in the grid modulo 25 (taking two rows at a time) gives PCNODDD DIGITSEVB HIGHLIGHTED KNISHIKADOH. With the odd digits highlighted, we may recognise the form of one of the creatures from the arcade game Space Invaders. A little research will reveal that it was designed by Tomohiro Nishikado.