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 14 |
asked | How to compute large modulos with pen and paper? |
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Nov 14 |
accepted | Fastest way to compute [1234567890]_200 with pen and paper |
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Nov 14 |
awarded | Commentator |
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Nov 14 |
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. |
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Nov 14 |
awarded | Editor |
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Nov 14 |
revised |
Fastest way to compute [1234567890]_200 with pen and paper added 2 characters in body |
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Nov 14 |
comment |
Fastest way to compute [1234567890]_200 with pen and paper Right. Thanks ! That was the simplification I was looking for. |
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Nov 14 |
asked | Fastest way to compute [1234567890]_200 with pen and paper |
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May 7 |
accepted | Proof of a Binomial Identity using a combinatorial argument |
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May 7 |
comment |
Proof of a Binomial Identity using a combinatorial argument Thanks for the help. This cleared things up ! |
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May 7 |
comment |
Proof of a Binomial Identity using a combinatorial argument Alright. Thanks. |
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May 7 |
comment |
Proof of a Binomial Identity using a combinatorial argument $(k+l) \times (k-l) = k^2 - l^2$ |
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May 7 |
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. |
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May 7 |
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 |
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May 7 |
accepted | Inductive Proof of a countable set Cartesian product |
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May 7 |
asked | Proof of a Binomial Identity using a combinatorial argument |
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May 2 |
awarded | Supporter |
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May 2 |
comment |
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 2 |
awarded | Scholar |
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May 2 |
accepted | Induction Proof and divisibility by $2^n$ |
