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I have $N+1$ numbers. If I sum any $N$ of them, the sum is divisible by $N$.
If one of the numbers is divisible by $N$, must all of the other numbers also be divisible by $N?$
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If a periodic non-zero function $f(x)$ exists such that $\sqrt{3}f(x) = f(x - 1) + f(x + 1),$ which of these options could be its fundamental period?
Note: A function $g(x)$ is periodic if there exists some $T$ such that $g(x) = g(x + T)$ for all $x.$ The fundamental period is the minimum positive $T$ such that this property holds.
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$\LARGE n¡=n^{{(n-1)}^{{(n-2)}^{.^{.^{.^{2^{\small1}}}}}}}$
If $n¡$ is defined as above, what are the last two digits of the number $2018¡?$
Note: $\large ¡$ is the factorial notation $!$ turned upside down. Note that $\large ¡$ keeps exponentiating while $!$ keeps multiplying.
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$\sum_{n=1}^\infty \binom{4n}{2n}^{-1}=X+\frac{\pi}{\sqrt{Y}}-\frac{2}{Z\sqrt{Z}}\ln\left(\frac{1+\sqrt{Z}}{2}\right)$
The equation above holds true for rational numbers $X$, $Y$, and $Z$. Find $\sqrt{XYZ}$.
Note: $\binom{\cdot}{\cdot}$ is the binomial coefficient. The first few terms of the series are as follows:
$\begin{aligned} \sum_{n=1}^\infty \binom{4n}{2n}^{-1} &= \frac{1}{\binom{4}{2}} + \frac{1}{\binom{8}{4}} + \frac{1}{\binom{12}{6}} + \cdots \\ &= \frac{1}{6}+\frac{1}{70}+\frac{1}{924}+\cdots. \end{aligned}$
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Let $A \subset \mathbb Z$ be a non-trivial set of integers $($i.e. $A \not= \emptyset$ and $A \not= \mathbb Z).$ We call such a set $A$ a "remarkable set of type $N$" if it has the following properties:
If $a$ is an element of $A,$ then $-a$ is also an element of $A$.
If $a$ is an element of $A$, then $a+N$ is also an elements of $A$.
If $a,b$ are elements of $A$ (not necessarily different), then $a + 2b$ is also an element of $A$.
How many (non-trivial) remarkable sets of type 18 are there?
Bonus: Generalize. If $N$ = $2^{e_2}\cdot 3^{e_3}\cdot 5^{e_5}\cdots$ is the prime factorization of $N$, how many non-trivial remarkable sets of type $N$ exist? What do they look like?
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