Let $E$ be the elliptic curve over $\mathbb{F}_3$ in medium Weierstrass form $E:y^2=x^3+x^2+x+1$. How to compute the number of points $|E(\mathbb{F}_{3^k})|$? I read that there are some formulas for computing number of points for short Weierstrass form by Frobenius endomorphism. But they don't wo...

Three fair coins are tossed, and we let $X_1$, denote the number of heads that appear. Those coins that were heads on the first trial (there were $X_1$, of them) we pick up and toss again, and now we let $X_2$, be the total number of tails, including those left from the first toss. We toss again ...

Should one approach by coordinates or by euclidean geometry? By pure geometry, I am not able to solve.

I was computing the homology of $S^3-\coprod_{i=1}^4 I_i$, where $I_i=[0,1]$ for all $i$ (they are being identified with an embedding). Intuitively, this should be homotopy equivalent to $S^1$, since removing one interval gives something homotopic to $\mathbb{R}^3$, removing another one gives an ...

How many complex number $z$ are there such that $|z+1|= |z+i|$ and $|z|=5$? My attempt : I got $2$, that is $ z=-2, z= +2$ , $|z| = {\sqrt{ 2^2+1}}$, $|z| = {\sqrt{(-2^2) +1}}$ Is it true ?

$17!$ is equal to $$35568x428096y00$$ Both x and y, are digits. Find x,y. So, $$17!=2^{15}*3^6*5^3*7^2*11*13*17=(2^3*5^3)*2^{12}*3^6*7^2*11*13*17$$ If there`s a product of $(2*5)^3$ Then this number has $3$ zeros at the end, so $y=0$ How do I find the $x$ now?