Using binomial distribution it's $\frac{\binom{5}{4}}{2^5}$
We want:
$$\log \binom{10}{0}\binom{10}{1}\cdots \binom{10}{10}=\log \binom{10}{0}^2\binom{10}{1}^2\binom{10}{2}^2\binom{10}{3}^2\binom{10}{4}^2\binom{10}{5}=\log (1\cdot 10^2\cdot 45^2\cdot 120^2\cdot 210^2\cdot 252)=\log(2^12\cdot 3^10\cdot 5^8\cdot 7^3)\approx 16.51$$
The question is wrong, no integer solutions exist to $2^n\equiv 5\pmod{15}$.
$2x+5y$ isn't a line.
I'll solve for $a,b,c$?
Multiplying them all together, $a^4b^4c^4=2^2\cdot 3^4\cdot 7^2\implies a^2b^2c^2=2\cdot 9\cdot 7$. So $c^2=\frac{a^2b^2c^2}{a^2b^2}=\frac{9}{2}$, now plug into eqn 3 to find a and then plug into eqn 1 to find b.
$a\cdot 2^r=1$, $a\cdot 32^r=16$. Dividing the two, $\frac{32^r}{2^r}=16^r=16\implies r=1$.
$\omega$ is a cube root of unity so write it in exponential form everywhere. This problem is easy to guess as well.
$(1111111+909091)(1111111-909091)=(2\cdot 10^6+2\cdot 10^4+2\cdot 10^2+2\cdot 10^0)(2\cdot 10^5+2\cdot 10^3+2\cdot 10)=4(10^6+10^4+10^2+10^0)(10^5+10^3+10)=4(10^{11}+10^{9}+10^{7}+10^9+10^7+10^5+10^7+10^5+10^3+10^5+10^3+10)=4(10^{11}+2\cdot 10^9+2\cdot 10^7+2\cdot 10^5+2\cdot 10^3+10)=4(102020202010)=408080808040$.
Don't be scared.
It's $1+i\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...\right)=1+i$.
$x^2-3x-80=0$. You know that $r=-\frac{3}{2}$ so to find $s$ subtract $\left(x-\frac{3}{2}\right)^2$ from the LHS.
