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Don't have much time these days...


Aug
12
awarded  Good Answer
Aug
7
awarded  Good Answer
Aug
7
awarded  Enlightened
Aug
7
awarded  Nice Answer
Aug
6
awarded  binomial-coefficients
Aug
5
answered Truncated alternating binomial sum
Aug
5
revised Elementary proof there are infinitely many primes of the form $4n+1$
edited tags
Jul
31
revised How prove $7\mid abcd$, if $a^{14}+b^{14}+c^{14}=d^{14}$
edited tags
Jul
30
awarded  Guru
Jul
28
comment How prove $\sqrt{r^2+c^2}$ is irrational
Tagging this is elementary-number-theory until proven otherwise.
Jul
28
revised How prove $\sqrt{r^2+c^2}$ is irrational
edited tags
Jul
27
comment Find the pair of values $a[i]$, $a[j]$ such that $a[i]\,\&\,a[j]$ is maximum
@900sit-upsaday: Thanks, I have cast my dupe vote there.
Jul
27
comment Find the pair of values $a[i]$, $a[j]$ such that $a[i]\,\&\,a[j]$ is maximum
You haven't clarified the bitwise AND part (this is a mathematics site :-)). Also, as written, the question gives the impression that you haven't put any effort into it, which clearly isn't true. Why not edit the question and put in some motivation and your ideas and requirement of a linearithmic algorithm, and turn this into good question (according to the standards of this site)?
Jul
27
revised Find the pair of values $a[i]$, $a[j]$ such that $a[i]\,\&\,a[j]$ is maximum
edited tags
Jul
27
comment Find the pair of values $a[i]$, $a[j]$ such that $a[i]\,\&\,a[j]$ is maximum
$a[i] \& a[j]$ is bitwise AND of $a[i]$ and $a[j]$? Where did you come across this problem? What is wrong with the trivial $\Theta(n^2)$ brute force algorithm? Is this a puzzle you are posing here?
Jul
26
revised Finding whether a sum of numbers in a set generate another number
edited tags
Jul
26
revised How to show $n(n+1)(2n+1) \equiv 0 \pmod 6$?
edited tags
Jul
25
revised Proving something about the Game Nim
edited tags
Jul
25
comment Asymptotic Behaviour Of A Bizarre Function 2
Related: math.stackexchange.com/questions/115824/…
Jul
25
comment Asymptotic Behaviour Of A Bizarre Function 2
@900sit-upsaday: Thanks! Wasn't aware of this nice feature. Been away from stackexchange long enough...