## Keep Your Distance by Pandiculator

### 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.