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Don't have much time these days...


Jul
27
revised Find the pair of values $a[i]$, $a[j]$ such that $a[i]\,\&\,a[j]$ is maximum
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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
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Jul
26
revised How to show $n(n+1)(2n+1) \equiv 0 \pmod 6$?
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Jul
25
revised Proving something about the Game Nim
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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...
Jul
24
comment How prove this $\sum_{cyc}\frac{x+y-2z}{(x+y)^2+z^2}=0$
Why is this tagged inequality?
Jul
24
comment Polynomial representation
@Minu: Yes, that is correct.
Jul
24
comment Polynomial representation
@Mathmo123: Don't know. Let's see Minu's response. (You might be right though)
Jul
24
answered Polynomial representation
Jul
24
revised Showing that if $p$ is prime, then $(p^4 + 4)$ can't be prime
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Jul
24
comment Prove that an expression is zero for all sets of distinct $a_1, \dotsc, a_n\in\mathbb{C}$
@DamianPavlyshyn: Yes, or more simply, replace $a_1$ by $z$, do some algebra to get a polynomial in $z$ which has infinite roots (any $z \ne a_i$), allowing setting $z=0$.
Jul
23
revised Prove that an expression is zero for all sets of distinct $a_1, \dotsc, a_n\in\mathbb{C}$
added 54 characters in body
Jul
23
answered Prove that an expression is zero for all sets of distinct $a_1, \dotsc, a_n\in\mathbb{C}$
Jul
23
comment Prove that an expression is zero for all sets of distinct $a_1, \dotsc, a_n\in\mathbb{C}$
Look at the Lagrange polynomial of $P(a_i) = a_i$
Jul
22
revised Discrete Mathematics Function Proof
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Jul
19
comment Proving that one of $a(1-b), b(1-c), c(1-a) \le \frac{1}{4}$
For $a,b,c \le 1$, see my comment to DanZimm. There is no need of $c$ if $a \le b$. There is an implicity renaming of variables going on. For instance if you chose $a=0.3, b = 0.1, c = 0.4$, we kind of have implicitly swapped $b$ and $c$ in our proof...
Jul
19
comment Proving that one of $a(1-b), b(1-c), c(1-a) \le \frac{1}{4}$
@DanZimm: If $c \ge 1$, then $b(1-c) \le 0$.
Jul
19
revised How could I improve this approximation?
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