Rom

you're right

delete my answer if you can

Choosing 2 adjacent vertices gets us 2 squares.  One is the original square ABCD.

There are 4 ways to choose adjacent vertices and this gets us a total of 5 unique squares.

Choosing diagonal vertices gets 2 squares each.  There are 2 pairs of diagonal vertices.

This gets us a total of 2x2=4 unique squares.

Thus there are a total of 5 + 4 = 9 unique squares, one of which is the original ABCD

I think it was probably meant to be asked has how many integers

$1 \leq n \leq 1000$ does the decimal representation of $\dfrac 1 n$ terminate.

I suppose one trick you could use is

$\lim \limits_{x\to -\infty}~f(x) = \lim \limits_{x \to \infty}~f(-x)$

$9\sqrt{56x^7y^{12}}= \\ 9\sqrt{4 x^6 y^{12}\cdot 14 x} = \\ 9(2x^3y^6)\sqrt{14x} = \\ 18x^3y^6\sqrt{14x}$

The problem is symmetric in x vs. 1/x so we can find the number of integers between 0 and 1000 that are either a power of two or 3 and just double the answer.

Note that any non-zero power of 2 is even and any non-zero power of 3 is odd so these two sets are disjoint.

There are 10 powers of 2, 0-9, and 6 non-zero powers of 3 that are less than 100.

We use non-zero power for 3 since we only want to count 1 = 2⁰ = 3⁰ once

That gets us 16 from 0 to 1000.  Doubling this we get 32 but we don't want to count $1 = \dfrac{1}{1}$

twice so we subtract 1 from this getting 31 as you found.

$\text{apparently }\sum \limits_{k=1}^n ~k! \cdot k = (n+1)! - 1 \\ 0!=1 \\ \text{so the expression shown }=(n+1)!-1+1 = (n+1)! \\ \text{so the largest prime }<(n+1) \text{ is what we're after}\\ \text{and this is }47.$

$\text{let }U \text{ be the set of whole numbers }1-100 \\ \text{and let }V \text{ be }U \text{ with the multiples of 3 and 4 removed}\\ |V| = |U| - |\text{multiples of 3 < 100}| - |\text{multiples of }4 \leq 100| + |\text{multiples of both 3 and 4} \leq 100|$

$|U| = 100 \\ |\text{multiples of 3}<100| = 33 \\ |\text{multiples of 4}\leq 100| = 25 \\ |\text{multiples of 3 and 4}\leq 100| = |\text{multiples of 12}\leq 100| = 8 \\ |V| = 100 - 33 - 25 + 8 = 50$

$z = r e^{i \theta},~0 \leq r, ~0 \leq \theta < 2\pi \\ z^4 = r^4 e^{i 4 \theta} = |z| = r = re^{i 2k\pi}, k \in \mathbb{Z} \\ r = 1,~ 4\theta = 2k \pi \\ \theta = 0,~\dfrac{\pi}{2},~\pi,~\dfrac{3\pi}{2} \text{ the solutions repeat after this} \\ z = 1, i, -1, -i$