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visits member for 2 years, 7 months
seen Jun 20 '13 at 18:41

When not programming I spend much of my free time avoiding the task of writing bios.


Jun
19
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.
Jun
19
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 :)
Nov
29
comment Modular Arithmetic Equations
Got it ! Thanks :)
Nov
29
comment Why is $x^7 = x$ true for every $x$ in $x ∈ Z/14Z$?
Why does it work for 2 and 7 too ?
Nov
29
comment Why is $x^7 = x$ true for every $x$ in $x ∈ Z/14Z$?
I know. But still don't see how this applies.
Nov
29
comment Why is $x^7 = x$ true for every $x$ in $x ∈ Z/14Z$?
I'm not sure where this is heading.
Nov
29
comment Modular Arithmetic Equations
CRT is probably the way to go but not sure how to apply it here.
Nov
15
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.
Nov
15
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.
Nov
15
comment How to compute large modulos with pen and paper?
Yeah, any power of 3 greater than will make the remainder 0.
Nov
15
comment How to compute large modulos with pen and paper?
Oh right ! Thanks
Nov
15
comment How to compute large modulos with pen and paper?
I don't see what number is multiple of 3 in the first case.
Nov
15
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.
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.
Nov
14
comment Fastest way to compute [1234567890]_200 with pen and paper
Right. Thanks ! That was the simplification I was looking for.
May
7
comment Proof of a Binomial Identity using a combinatorial argument
Thanks for the help. This cleared things up !
May
7
comment Proof of a Binomial Identity using a combinatorial argument
Alright. Thanks.
May
7
comment Proof of a Binomial Identity using a combinatorial argument
$(k+l) \times (k-l) = k^2 - l^2$
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.
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