Rom

looks good

There's no number theoretic solution to this that I know of.

You just have to go through the list and check each one.

By hand this is absurdly tedious.  Something you'd give a team of 12th century monks a month to complete.

In software it's pretty trivial to code up a quick program to dump the list of them and it's length.

How many monks do you have available?

The hint given is wrong.  You need to group the terms as follows.

\(x^3 + 7x^2 -6x-42 = \\ (x^3 -6x) + (7x^2 - 42) = \\ x(x^2 - 6) + 7(x^2 - 6) = \\ (x+7)(x^2-6)\)

Go through the entire test answering those questions that you can easily (or that believe you can easily) answer first.

Only then come back and attack problems you don't immediately see the answer to.

If you find yourself really getting bogged down in a problem move on.

Basically save the hardest problems for last.

How do you know which ones are the hardest problems?  Well that's the trick isn't it.

\(\text{A quadratic has a common root when it's discriminant is 0}\\ p(x) = a x^2 + bx+c \Rightarrow D = b^2 - 4ac\\ D = b^2 - (4)(1988)(8891) = 0\\ b^2 = 70701232\\ b = \pm 4 \sqrt{4418827}\)

\(6 = \dfrac{35}{x} - \dfrac{49}{x^2}\\ 6x^2 = 35x - 49\\ 6x^2 - 35x +49 = 0\\~\\ x = \dfrac{35\pm \sqrt{35^2 - (4)(6)(49)}}{12} = \left\{\dfrac 7 3, \dfrac 7 2\right\} \)

\(\text{The key is that}\\ a^n \mod p = (a \mod p)^n \mod p\)

\(13^{13}+5 \mod 6 = \\ 1^{13} + 5 \mod 6 = \\ 6 \mod 6 = 0\)

I'll give you a thumbs up but you butchered that joke.

\(|2x+2|\geq 2\\~\\ 2x+2\geq 2 \Rightarrow x\geq 1\\ 2x+2 \leq -2 \Rightarrow x \leq -2\\~\\ S = \{x : x \in (-\infty, -2] \cup [1,\infty)\}\)

I leave the second one to you.

\(\text{Notice the expression is symmetrical in $x$ and $y$}\\ \text{This suggests that any solution will be such that $x=y$}\\ $\text{Letting $x=y$ we have}\\ 3x^4 - 17x^2 + 20 = 0\\ \text{let $u=x^2$}\\ 3u^2 - 17u + 20 = 0\\ \text{Apply the quadratic formula}\\ u = \dfrac{17 \pm \sqrt{17^2 - (4)(3)(20)}}{6} = \left(\dfrac 5 3,~4\right)\\ \text{you want integer solutions so we can ignore the first one}\\ x^2 = 4\\ x=\pm 2\)

\(\text{So the max of $x+y$ is $2+2=4$}\)