Project Euler time! I'm working on https://projecteuler.net/problem=32 if anyone wants to join me.

Question

anonymous · Answer

Sure :) What programming language do you use?

kainui · Answer

I'm using Java at the moment, but I use MatLab sometimes. What about you?

kainui · Answer

It appears that for our numbers a*b=c that we can set up this inequality $$\log_{10}a+\log_{10}b \ge \log_{10}c$$ and we also know $$\lfloor \log_{10} a \rfloor +\lfloor \log_{10}b \rfloor +\lfloor \log_{10}c  \rfloor =10$$ since we need all of the digits, and in addition to that we also know that the sum of digits is: 45.

So that kind of gives us a couple things to think about.

anonymous · Answer

Sorry, had this typed out but never actually pressed send. I'm using C# :) I can't for the life of me think how to find these without doing a huge brute force - either by generating all 362880 permutations of 1-9 then factorising them and checking the factors, or trying loads of factors and count the number of pandigital numbers I find. Either of those approaches will take a long, long time. I'll have a think and get back to you if I find anything.

kainui · Answer

Well that's what this is for, so we can figure it out and brainstorm together.

kainui · Answer

I think it's probably a good idea to start at the factors since we will know a couple of things. We know that their product will have to have only so many digits, so we won't try multiplying something like 1234*567 since obviously there are too many digits to create a 2 digit number.

anonymous · Answer

It's just gone 3:40am here so I've got to go to bed, but I'll continue with this in the morning.