network profile
| bio | website | |
|---|---|---|
| location | ||
| age | 21 | |
| visits | member for | 1 year |
| seen | May 13 at 22:16 | |
| stats | profile views | 25 |
I'm a dev bi***
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Nov 29 |
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Modular Arithmetic Equations Got it ! Thanks :) |
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Nov 29 |
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Why is $x^7 = x$ true for every $x$ in $x ∈ Z/14Z$? Why does it work for 2 and 7 too ? |
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Nov 29 |
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Why is $x^7 = x$ true for every $x$ in $x ∈ Z/14Z$? I know. But still don't see how this applies. |
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Nov 29 |
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Why is $x^7 = x$ true for every $x$ in $x ∈ Z/14Z$? I'm not sure where this is heading. |
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Nov 29 |
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Modular Arithmetic Equations CRT is probably the way to go but not sure how to apply it here. |
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Nov 15 |
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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. |
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Nov 15 |
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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. |
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Nov 15 |
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How to compute large modulos with pen and paper? Yeah, any power of 3 greater than will make the remainder 0. |
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Nov 15 |
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How to compute large modulos with pen and paper? Oh right ! Thanks |
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Nov 15 |
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How to compute large modulos with pen and paper? I don't see what number is multiple of 3 in the first case. |
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Nov 15 |
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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. |
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Nov 14 |
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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. |
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Nov 14 |
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Fastest way to compute [1234567890]_200 with pen and paper Right. Thanks ! That was the simplification I was looking for. |
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May 7 |
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Proof of a Binomial Identity using a combinatorial argument Thanks for the help. This cleared things up ! |
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May 7 |
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Proof of a Binomial Identity using a combinatorial argument Alright. Thanks. |
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May 7 |
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Proof of a Binomial Identity using a combinatorial argument $(k+l) \times (k-l) = k^2 - l^2$ |
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May 7 |
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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. |
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May 7 |
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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 |
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May 2 |
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Induction Proof and divisibility by $2^n$ Thanks. Excellent answer. Is there a way to get a closed recursive formula that defines A(n + 1) in terms of A(n) ? |
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May 1 |
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Inductive Proof of a countable set Cartesian product I made it with your help. Many thanks ! |
