Enigma 1760

Spoiler for New Scientist Enigma 1759: “Squares and cubes” (Follow the link to see the puzzle.)

The number which is both a square and a cube must be a sixth power: 26 = 82 = 43 = 64, or 36 = 272 = 93 = 729.

Note that the only cubes with 2 or 3 digits are 27, 64, 125, 216, 343, 512, and 729.

Try 64 at 6 down. Then 6 across must be a square: 625 or 676. If 625 then 5 down has middle digit 5 which means it’s 256. Then 7 across can only be 64, which duplicates. If 6 across is 676 then there is no possibility for 2 down. So 6 down is not 64.

Try 64 at 3 down. Then 4 across is 144, 324, 484, or 784. If 144 then 2 down must be 216 or 512. The former leaves nothing for 1 across; the latter forces 1 across to be 25. 5 down must be 441 or 484 (400 is ruled out because we don’t allow a leading zero for 7 across), but no 3-digit cube ends with 8 so it’d have to be 441. Then 7 across must be 16, 6 across is 324, and 6 down is 36. If 4 across is 324 the middle digit of 2 down is 3, but there are no 3-digit squares or cubes with 3 as the middle digit. If 4 across is 484 or 784 then 5 down is 841 and 7 across is 16. In the 484 case 2 down must be 144 or 441 or 841, none of which leaves anything for 6 across. In the 784 case 2 down must be 576 or 676 either of which leaves nothing for 6 across.

The other possibilities are 729 at 2 down or 5 down. 5 down doesn’t work because it leaves nothing for 7 across.

With 729 at 2 down, 6 across can only be 196, with 6 down being 16. Then 5 down must be 169, 361, or 961. The first leaves nothing for 7 across while the latter two leave only 16, which duplicates.

So there is indeed only one solution:

2 5   6
  1 4 4
3 2 4  
6   1 6
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