BrainOverfl0w
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 Nov15 comment How to compute large modulos with pen and paper? Yeah, any power of 3 greater than will make the remainder 0. Nov15 accepted How to compute large modulos with pen and paper? Nov15 comment How to compute large modulos with pen and paper? Oh right ! Thanks Nov15 revised How to compute large modulos with pen and paper? edited body Nov15 comment How to compute large modulos with pen and paper? I don't see what number is multiple of 3 in the first case. Nov15 comment How to compute large modulos with pen and paper? I notice the pattern for the first one. But I can't prove why it's accurate to predict so. Nov14 asked How to compute large modulos with pen and paper? Nov14 accepted Fastest way to compute [1234567890]_200 with pen and paper Nov14 awarded Commentator Nov14 comment Fastest way to compute [1234567890]_200 with pen and paper Sorry about that. Seems like the previous notation wasn't an international standard. But it was $[1234567890]_200$ which is the number in base 200 and hence similar once reduced to the modulo. Nov14 awarded Editor Nov14 revised Fastest way to compute [1234567890]_200 with pen and paper added 2 characters in body Nov14 comment Fastest way to compute [1234567890]_200 with pen and paper Right. Thanks ! That was the simplification I was looking for. Nov14 asked Fastest way to compute [1234567890]_200 with pen and paper May7 accepted Proof of a Binomial Identity using a combinatorial argument May7 comment Proof of a Binomial Identity using a combinatorial argument Thanks for the help. This cleared things up ! May7 comment Proof of a Binomial Identity using a combinatorial argument Alright. Thanks. May7 comment Proof of a Binomial Identity using a combinatorial argument $(k+l) \times (k-l) = k^2 - l^2$ May7 comment Proof of a Binomial Identity using a combinatorial argument I see how you could use this as an algebraic proof but I don't think this can be used to prove it using a combinatorial argument. May7 comment Proof of a Binomial Identity using a combinatorial argument I don't see exactly where this is heading. I can agree on the fact of splitting the initial $2k$ items group into two smaller ones but I don't get were the $k^2$ and $l^2$ are coming from