## Square Time Sums by Oyler

### Puzzle explanation

The highlighted rows and columns are square numbers containing all of the digits 1 to 9.

Though it wasn’t required for solving the puzzle, the four cells enclosed by the highlighted lines contained the digits {1, 4, 7, 9}. These are the only possible values that result from successively adding up the digits of a square number until a single digit is reached. For example, the sum of the digits of 289 is 19, the sum of the digits of 19 is 10, and the sum of the digits of 10 is 1. For the highlighted square numbers, the result is always 9 because the digits 1 to 9 add up to 45.

### Puzzle solution process

This is one possible solution path. The squares in the triples can’t contain 0 or a repeated digit; the only ones with two digits are {16, 25, 36, 49, 64, 81} and the only three-digit examples are {169, 196, 256, 289, 324, 361, 529, 576, 625, 729, 784, 841, 961}.

All 12 of the triples’ squares appear as grid entries. In particular, (J, K, L) are (1ac, 9ac, 15ac), which means there’s no repeated digit in 1dn = 7X. From 18ac we know that A > X, so X isn’t 81; of the five other possible values for X, four give a repeated digit in 7X, which leaves 29ac = X = 25 and 1dn = 175. Then the only square that fits at 1ac is J = 16. As 9ac (K) starts with 7, it must be one of {729, 784}, and 15ac (L, starting with 5) must be a permutation of {5, 3, 4, 8} or {5, 2, 3, 9} respectively. As a square, L must end with 4 or 9, so it’s in the range [5239, 5834] and its square root is between approximately 72.4 and 76.4; the only integer in that range whose square ends with 4 or 9 is 73, so 15ac = L = 5329 and 9ac = K = 784. This completes 2dn = 683, which is prime as expected.

As 44dn = 5 × 19dn has two digits, 19dn must be < 20, so 18ac = A − 25 ends with 1, which means A ends with 6, forcing 49ac = A = 36 (as 16 is already taken), which gives 18ac = 11 and 20ac = 252, completing 10dn = 422.

As 43ac is a prime, it ends in one of {1, 3, 7, 9}, which is also the first digit of 44dn = 5 × 19dn, so 44dn must be one of {70, 75, 90, 95} with 19dn being one of {14, 15, 18, 19}. From the grid, 40dn = 19dn + Y ends with 6, and as Y is a square (ending in one of {1, 4, 5, 6, 9}), 19dn ends with one of {0, 1, 2, 5, 7}. The only value that satisfies both constraints is 19dn = 15, making 44dn = 75.

Now 25ac (Q) starts with 5, so it’s one of {529, 576}. That gives {59, 56} respectively for 21dn = 48ac − 36, so 48ac is one of {95, 92}. From the clue, 16dn is even, so 37ac is even, and as 48ac is a multiple of 37ac it must also be even, so 48ac = 92, making 21dn = 56 and 25ac = Q = 576. As 92 = 2×2×23, its only two-digit factors are {23, 46, 92}, so 37ac = 46 (even, and not the same as 48ac). That gives 4 for the middle digit of 32dn (Y), and the only suitable square is 32dn = Y = 841 (completing 43ac = 17, prime as expected), which makes 45ac = 795 and 40dn = 856.

Now 39dn is _93, a multiple of 43ac = 17, so the multiplier (_93/17) must end with 9. The only one that fits is 29, giving 39dn = 493.

From the grid, 10dn = 422. As 33ac is the sum of two two-digit numbers, it’s in the range [101, 199], so 38dn = 10dn − 33ac is in the range [223, 321]. From the grid, 38ac is then either 2_48 or 3_48 and it’s a multiple of 2A + 2J + 3X = 179. The multiplier must end in 2 and it’s between 2048 / 179 ≈ 11.4 and 3948 / 179 ≈ 22.1, so it’s one of {12, 22}; 22 × 179 = 3938 doesn’t fit, so 38ac = 2148 (12 × 179). That limits 38dn to {229, 239, ..., 299} and 33ac to {123, 133, ..., 193} and makes 34dn = 317 (prime as expected).

From 12dn we know that B > Q, and B can’t contain 3 or 6 (because A is 36), so it’s one of {729, 784, 841}. As 784 and 841 are already used, 17dn = B = 729, which makes 12dn = 153 (completing 22ac = 32). That means 16ac = P − 36 ends in 5, so P ends in 1, ie 11ac = P = 81, which makes 16ac = 45, 27ac = 898, 4dn = 829, 7dn = 35 and 14dn = 1273.

From the grid, 27dn and 28dn both start with 8, so 33ac = 27dn + 28dn is in the range [161, 177]; from the grid it ends with 3, so 33ac is one of {163, 173}. If 33ac were 163 then 27dn would be 86 and 28dn would be 163 − 86 = 77, which doesn’t fit; therefore 33ac = 173, 27dn = 87 and 28dn = 86, which also gives 38dn = 249.

We have A = 36 and B = 729, so 30dn = C uses the digits {1, 4, 5, 8} and from the grid it starts with 5, so as a square it ends in one of {1, 4}; 46ac ends with the same digit and starts with 8 (the last digit of 31ac) and can’t be 81, which is already used, so 46ac = 84 and 31ac = 48. C is one of {5184, 5814} but 31ac = 48 isn’t a factor of 5814, so 30dn = C = 5184. As 26dn = 17ac − 31ac and 17ac ends with 2 (from the grid), 26dn ends with 4, ie 26dn = 744, which makes 17ac = 792. Then 6ac = 63 and 3ac = 98.

We have X = 25 and Y = 841, so 8dn = Z is a permutation of {3, 6, 7, 9}. From the grid we know the third digit is 9, so to be a square it has to end in 6 and it’s one of {3796, 7396} (completing 23ac = 67, prime as expected). That makes 13ac, which is a multiple of 75 − 16 = 59, one of {531, 571}; 571/59 isn’t an integer, so 13ac = 531 and 8dn = Z = 7396.

We have P = 81 and Q = 576, so 24dn = R is a permutation of {2, 3, 4, 9}. From the grid we have _2__, and to be a square it must end in 4 or 9, so R is one of {3249, 3294, 4239, 9234}. From the clue for 16dn, R = 16dn/2 − C − L − Z, and from the grid 16dn is 42_16, so the maximum for R is 42916/2 − 5184 − 5329 − 7396 = 3549, which limits R to {3249, 3294}. Its square root is between 50 and 60 (as R is between 2500 and 3600) and must end in one of {2, 3, 7, 8}; the only one that works is 57, giving 24dn = R = 3249, which makes 16dn = 42316.

From the grid we now have 36ac = 414, which factorises as 2×3×3×23, so the only factor that fits 6_ for 35ac is 35ac = 69.

From the grid, the prime at 41ac is now 98_; the last digit can’t be 1 or 7 because they would make it a multiple of 3, so the prime at 42dn starts with 3 or 9. But 8ac = 7_ is a multiple of 42dn, so the first digit can only be 3 (with 41ac = 983) and the possible primes are {31, 37}. It can’t be 31 because that has no multiples in the 70s, so 42dn = 37 and 8ac = 74.

47ac is a factor of 42316, matching 7_. It can’t be 70 or 75 (which don’t have multiples ending in 6) or 74 (already used at 8ac); of the rest, only 71 divides evenly, so 47ac = 71.

From the grid 5dn is _8; it’s a factor of both 17ac = 792 = 2×2×2×3×3×11 and 30dn = 5184 = 2×2×2×2×2×2×3×3×3×3, so it must be some combination of the factors of 2×2×2×3×3, ie, a factor of 72. The only matching value is 5dn = 18, which completes the grid.

Of the grid’s rows and columns, only the 6th and 9th in each direction consist of the digits 1 to 9 once each. They are also the only rows and columns that are nine-digit squares, so they’re doubly thematic.