thedudemanguyperson

The LHS is an integer,  so the RHS must be as well and conversely $i=n^2$ for integer $n$.

Then sum of divisors of $n^2$ is $n^2+n+1$ then use the divisor sum formula.

That's 100% correct

You can just experiment. For $3$ dice, there are $\binom{3}{2}$ ways to arrange and $\frac{5}{6}\cdot \frac{5}{6}\cdot \frac{1}{6}$ probability. This yields $\frac{25\cdot 3}{216}=\frac{25}{72}$ which works. So $n=3$.

Note that $2^{13}\equiv 2^1\pmod{10}$ so $2^{n}=2^{n+12}\pmod{10}$. Then $2^{1000}\equiv 4\pmod{12}$ so $2^{2^{1000}}\equiv 2^4\equiv 6\pmod{10}$. The answer is $6+1=7$.

If they have four factors they can be of the form $p_1\cdot p_2$ or $p^3$ for all $p_1,p_2,p$ prime.

Let's do $p_1\cdot p_2$ first. We have $2\cdot 3=6, 2\cdot 5=10, 2\cdot 7=14, 2\cdot 11=22, $ etc. if $2$ is the first. Otherwise we can have $3\cdot 5=15, 3\cdot 7=21$, etc. So the smallest $5$ are $6,10,14,15,21$.

Then for $p^3$ we can have $2^3=8$ and the rest are larger than $21$. 

So the numbers are $6,8,10,14,15$. Sum them up.

We can't have $2\alpha\in (0,\pi/2]$ because then $\tan(2\alpha)>0$ and $k>0$ so $-\frac{1}{4k}<0$!!!! 

It's $\text{total number of ways} - \text{ways where no two tomatoes are together}$.

The first one is $8!$.

The second one is harder. We can have $TBTBTBTB$ or $BTBTBTBT$. The first one yields $(4!)^2$ as does the second.

The answer is then $8!-2(4!)^2=39168$.

$2x^2-6x+4$ is "in simplified form in terms of $x$".

You're done already!!!!

Right, others besides just $\frac{3\pi}{4}$ work. I was just giving an example, because the rest follow from the period $2\pi$. In fact, the general solution is $\theta=2\pi n\pm \frac{3\pi}{4}$ for integer $n$.