Reputation
31,326
Next tag badge:
96/100 score
28/20 answers
Badges
2 34 93
Newest
 Nice Answer
Impact
~599k people reached

7h
comment How to prove that a very large number is not prime
That is not a very large number. It is tiny compared to the numbers that can be routinely checked for primality these days.
1d
comment Find all functions $f(f(f(…(f(x_1,x_2),x_3),…),x_{2016}))=x_1+x_2+…+x_{2016}$
@Roman83: If you don't know the solution to this problem, it is absolutely unacceptable to change $2007$ to $2016$ like you did. You may have turned an easy question into an impossible one, which is just a waste of time for everybody.
1d
comment In tournament poker, can optimal players be targeted to force them into negative EV?
Game theory does not dictate that there is an optimal strategy in multi-player poker, precisely because of the possibility of collusion (which you so clearly demonstrate in your post). If two players in a three-player game are in cahoots, they can almost certainly clean out the third player.
2d
comment On a circle, will a point moving in irrational steps ever land at a point it has been at previously?
@Jens: If you take the trouble to formulate that question more precisely, you will probably find that it answers itself.
May
1
comment Can we find prime numbers with any sum of digits (except those divisible by three)
@Farewell: Most mathematicians would prefer to specify $n > 1$, rather than arbitrarily declare $1$ to be prime. It's sort of like saying "for the purposes of this question, let us consider $1$ to be even". (But it's not quite that bad, because in the dim and distant past there were mathematicians who counted $1$ as prime.)
Apr
26
awarded  Nice Answer
Apr
25
comment Solve $y'=e^{x^2}y$ (with $3$ terms only) in using power series
$\int x^{2n}dx$ is $\dfrac{x^{2n+1}}{2n+1}$, not $\dfrac{x^{2n+1}}{n+1}$. I only spotted this because of the discrepancy with Christian Blatter's answer, which gives $e^c(1+x+\frac12 x^2+\frac12 x^3+\cdots)$.
Apr
25
comment Is there a way to cut a an ellipsoid with a plane such that it gives an circle?
@user202729: Yes, this is the best way to look at it. (But your axes should be $1/\sqrt 2, 1/2,$ and $1/\sqrt 3$ instead of 1/2, 1/4, and 1/3.)
Apr
23
comment $0^0$ is undefined, but sometimes defined as $1$?
@REr: But $\lim_{y\to 0_+}0^y = 0$. So it's not that simple.
Apr
21
answered ECB mode decryption
Apr
21
comment Sorting the digits of $\pi$
@bluehallu: If you insist on making yourself look silly in front of your prospective employees, that is your right. But if you post a meaningless question on a maths forum, you will get called on it.
Apr
21
comment Sorting the digits of $\pi$
@bluehallu: So you are deliberately asking them a nonsense question, to see how they handle it? That is an unusual approach. Especially when only the best and the worst will say "I'm sorry, I don't understand."
Apr
21
answered Sorting the digits of $\pi$
Apr
21
comment A set of positive integers {1, 2, 3, 4, 5, 6, 7, 8, 9} is used to form a pack of nine cards.
If the largest card is a $5$, you can list the possibilities. How many are there? And why is this equal to $^4C_3$ instead of your $^5C_4$?
Apr
21
comment Why is $ \{(1/2)^n : n \in \mathbb{N} \} \cup \{ 0 \} $ not compact?
Your second sentence is rather sloppy! The interval $(-2,2)$ also contains the sequence $(x_n)$ and its limit $0$, but it is not closed.
Apr
20
comment How to square both the sides of an equation?
@ArnaudD.: I can't make sense of that comment.
Apr
20
comment How to square both the sides of an equation?
@Leg: if $x+3$ is negative, then $\sqrt{x+3}$ is undefined. So it makes no sense to say $(x+3)^{3/2} = \sqrt{x+3}\cdot|x+3|$. You are just muddying the waters for the OP.
Apr
20
comment How to square both the sides of an equation?
@ArnaudD.: No, but you get a valid equation. $a = b \implies a^2=b^2$.
Apr
20
comment How to square both the sides of an equation?
@ArnaudD.: You can always square both sides.
Apr
20
comment if $g|ab$ , $g|cd$, $g|ac+bd$, show $g|ac$ and $g|bd$.
Adding your two inequalities gives $2n \le r+s+t+u$. And if $r+t=s+u$, this gives $2n \le 2(r+t)$, so $n \le r+t$; similarly $n \le s+u$.