## What a Turn-up! by IOA

### Puzzle explanation After deducing the top row as 66066655, 11 segments are removed to form CLUELESS, as 1ac is the only grid entry for which no clue was given. The positioning of digits (particularly 1s) within their cells was ignored in the marking.

### Puzzle solution process

This is one possible solution path. As the filled grid’s contents are symmetrical, the asymmetrical digits 3, 4 and 7 can’t appear in any cells. Of the only allowable digits, 6 and 9 rotate to form each other, while 0, 1, 2, 5 and 8 retain their value on rotation.

10ac is double a square; any n2 ends in one of the digits {0, 1, 4, 5, 6, 9} so 2n2 ends in one of {0, 2, 8}. The rotational symmetry means that the first digit of 18ac is the same as the last digit of 10ac (rotated), and as 18ac can’t start with a zero, 10ac has to end in 2 or 8. Then 6dn and 16dn are 3-digit squares with 2 or 8 as the middle digit. The squares that don’t contain 3, 4 or 7 are {100, 121, 169, 196, 225, 256, 289, 529, 625, 900, 961}, of which 289 isn’t a possibility because its rotation is 682, not a square, so the middle digit must be 2 and 6dn and 16dn are either {121, 121} or {529, 625}.

Now 10ac = 2n2 ends in 2, so its n must end in one of {1, 4, 6, 9} and be between √(102/2) ≈ 7.1 and √(992/2) ≈ 22.3, ie one of {9, 11, 14, 16, 19, 21}; they give values of {162, 242, 392, 512, 722, 882} respectively, of which only {162, 512, 882} consist of allowed digits. The corresponding values for 18ac are {291, 215, 288}; 291 is an obvious multiple of 3 (the sum of its digits is 12) and 291/3 = 97, a prime; 215 is an obvious multiple of 5 and 215/5 = 43, a prime; so the only one that isn’t a product of two primes is 18ac = 288, making 10ac = 882.

5dn is a string of Fibonacci numbers in ascending order and has 8 as the second digit. There are only 38 Fibonacci numbers of up to 8 digits and calculating them all is not onerous. The only ones that consist of allowed digits are {1, 2, 5, 8, 21, 55, 89, 610}, so the first number in 5dn is one of {1, 2, 5}. It can’t end with 610 because that would make 2dn start with 0, so the only way to fill up 5dn is _|8|21|55|89, which makes 2dn 6855128_, ending in {1, 2, 5}. The digit sum of 68551281 is 36, a square; the digit sum of 68551282 is 37, a prime; so 2dn = 68551285 (with a digit sum of 40) and 5dn = 58215589.

11ac, which can be split into two single digits and their 2-digit sum, now has 5 as the second digit, so the parts comprise either 5 + v = 1w or x + y = 15, where {v, w, x, y} are all single digits. The possibilities for the former are {5+5=10, 5+6=11, 5+7=12, 5+8=13, 5+9=14}, of which only {5+5=10, 5+6=11} consist of allowed digits; for the latter the possibilities are {6+9=15, 7+8=15}, with only 6+9=15 allowed. To fit the grid, the only possibilities are {6|5|11, 15|6|9, 15|9|6}; 6511 would give 1159 for 17ac, with a digit sum of 16, not a Fibonacci number, so 17ac is one of {6951, 9651} and 11ac is one of {1569, 1596}. Because 3dn is a palindrome, its opposite 9dn is also a palindrome; from 11ac we know 9dn’s second digit is 6 or 9, so its sixth digit is the same, and by rotation the second and sixth digits of 3dn are also 6 or 9.

That makes the square 7ac end in 6 or 9, so it’s the square of a number ending in one of {3, 4, 6, 7}; also we know its second digit is 8 and it starts with one of {1, 2, 5, 6, 8, 9}. If 7ac starts with 1 then its root is between √18116 ≈ 134.6 and √18999 ≈ 137.8, but 1362 = 18496 and 1372 = 18769 both contain disallowed digits; if it starts with 2 then its root is between √28116 ≈ 167.7 and √28999 ≈ 170.3, but there are no integers ending in {3, 4, 6, 7} in that range; if it starts with 5 then its root is between √58116 ≈ 241.1 and √58999 ≈ 242.9, but 242 doesn’t end with {3, 4, 6, 7}; if it starts with 6 then its root is between √68116 ≈ 261.0 and √68999 ≈ 262.7, but 262 doesn’t end with {3, 4, 6, 7}; if it starts with 8 then its root is between √88116 ≈ 296.8 and √88999 ≈ 298.3, but 2972 = 88209 would make 9dn start with 0; therefore 7ac starts with 9 and its root is between √98116 ≈ 313.2 and √98999 ≈ 314.6, ie it’s 314, so 7ac = 98596 and 19ac = 96586 (with a digit sum of 34, a Fibonacci number).

As palindromes, we can fill in more of 3dn as 66___66 and 9dn as 99___99, which resolves 11ac = 1569 and 17ac = 6951. Now 14dn is __16_ and as a square it ends in one of {1, 5, 6, 9}; we can rule out 5 because the square of any number ending in 5 ends in 25. Its opposite 1dn is then _91__, starting with one of {1, 6, 9}. If it starts with 1 then its root is between √19111 ≈ 138.2 and √19199 ≈ 138.6; if it starts with 9 then its root is between √99111 ≈ 314.8 and √99199 ≈ 314.96; in both cases there are no integers in the range, so 1dn starts with 691 and its root is between √69111 ≈ 262.9 and √69199 ≈ 263.1, ie 263, which makes 1dn = 69169 and 14dn = 69169.

For 8dn we have 56___8_ and it’s equal to 2n2+1 for some value of n. Any 2n2+1 ends in one of {1, 3, 9}, but 3 isn’t allowed in the grid; if the last digit is 1 then 2n2 ends in 80 and n2 ends in 40 or 90, but there are no such squares (for any n ending in 0, its square ends in 00); therefore 8dn ends in 89, 2n2 ends in 88, n2 ends in 44 or 94 and n ends in one of {2, 8}. The value of n is between √((5600089-1)/2) ≈ 1673.3 and √((5699989-1)/2) ≈ 1688.2, so it’s one of {1678, 1682, 1688}; the corresponding values for 8dn are {5631369, 5658249, 5698689}, of which the first two have disallowed digits, so 8dn = 5698689 and 4dn = 6898695 (which is divisible by 5 and none of its other digits).

For 9dn we have 99___99, so its digit sum is in the range [36, 63]; the only Fibonacci number in that range is 55, so the middle 3 digits must add up to 19. The only palindromic sequences of allowed digits that have a sum of 19 are 595 and 919. We can reject 595 because it would give 6595_816 for 13ac, which has 2 digits (the 5s) by which it’s not divisible, so 9dn = 9991999 and 3dn = 6661666, which completes 13ac = 65991816 and 15ac = 91816659.

16ac is _269 and the first digit is one of {1, 5, 6}; the only one that gives a prime digit sum is 16ac = 6269, making 12ac = 6929, 6dn = 529 and 16dn = 625.

20ac is now 5599__99; the three known 2-digit segments have a sum of 55+99+99 = 253 and the only cube in the range [253, 352] is 343, so the missing digits are 90, making 20ac = 55999099 and 1ac = 66066655. The grid is now complete.

In the style of a calculator display, removing segments from a 6 can produce {C, E, F, I, L, S} and perhaps an approximation of G; a 0 can only produce {I, L, O, U}; and a 5 can only produce {S} (with no removals). A little experimentation reveals SELFLESS and CLUELESS as possibilities from 66066655, of which the latter is appropriate, as no clue is given for 1ac.

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