### Random Post

### Recent Posts

### Recent Comments

Jim Randell on Enigma 402: A DIY puzzle | |

geoffrounce on Puzzle 71: All wrong, all… | |

Jim Randell on Puzzle 71: All wrong, all… | |

geoffrounce on Enigma 1246: Triangle squ… | |

Hugh Casement on Enigma 1110: Dots and lin… |

### Archives

### Categories

- article (11)
- enigma (1,079)
- misc (2)
- project euler (2)
- puzzle (21)
- site news (42)
- tantalizer (21)
- teaser (3)

### Site Stats

- 157,038 hits

I wrote a generic cross-figure solver for

Enigma 1755(which can also be used to solveEnigma 1740andEnigma 1730– all also set by Peter Chamberlain). Here it is used to solve this puzzle. It runs in 66ms.Solution:The answer to 6 across is 324. The answer to 7 across is 16.This is what the completed grid looks like:

I’ve added this code to my

enigma.pylibrary, so that it can be used to solve similar puzzles.Same approach before, it is easy to modify the previous code, but I like to change the algorithm this time,

a^2=16,25,36,49,64,81,100,121,144,169,196,225,256,289,324,361,400,441,484,529,576,625,676,729,784,841,900,961

b^3=27,64,125,216,343,512,729

64 is the only common number for a^2 and b^3

By trial and error we can solve the full grid.

729 is also both a square and a cube.

Jim thank you for your warning , very true , there are two common numbers the other being 729 as you say, however the solution lies in one of them only.