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The National Cipher Challenge

Reply To: Maths

#97975
USB-C_is_supreme
Participant

This seems like more of a brute force method @ByteInBits @Astralica What I thought is that for the current possibilites, 2 is the highest common factor for a! and b!, so you can factorise out 4: 4(…) If 3 is a common factor, then you can factor out 9, making it impossible for any other triples.

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