BrainOverfl0w
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 Jun19 comment Number of solutions for $\sum_{i=1}^{4} x_i < 22$ with condition. Thanks, just got confirmation that there is a mistake on the solution sheet. Jun19 comment Number of solutions for $\sum_{i=1}^{4} x_i < 22$ with condition. Ok. I guess the wrong answer is on the teacher's solution sheet. Thanks :) Nov29 comment Modular Arithmetic Equations Got it ! Thanks :) Nov29 comment Why is $x^7 = x$ true for every $x$ in $x ∈ Z/14Z$? Why does it work for 2 and 7 too ? Nov29 comment Why is $x^7 = x$ true for every $x$ in $x ∈ Z/14Z$? I know. But still don't see how this applies. Nov29 comment Why is $x^7 = x$ true for every $x$ in $x ∈ Z/14Z$? I'm not sure where this is heading. Nov29 comment Modular Arithmetic Equations CRT is probably the way to go but not sure how to apply it here. Nov15 comment How many numbers exists that are smaller than $p$ and prime with $p$? Thanks, in short $3947$ is the answer I was looking for. Euler's totient function clarified my confusion. Nov15 comment How many numbers exists that are smaller than $p$ and prime with $p$? @JonasKibelbek : I edited the question to be clearer. You're right that it's a different question. Nov15 comment How to compute large modulos with pen and paper? Yeah, any power of 3 greater than will make the remainder 0. Nov15 comment How to compute large modulos with pen and paper? Oh right ! Thanks 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 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 comment Fastest way to compute [1234567890]_200 with pen and paper Right. Thanks ! That was the simplification I was looking for. 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