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Modular arithmetics as a way of calculating modulo some number $n$ is a powerful tool. However, it can be made simpler, and thus even more effective.

First, recall that numbers having the same remainder when divided by $n$ behave in exactly the same way in such a calculation and yield exactly the same results. $8 * 4 + 11 \equiv 3*(-1) + 6 \equiv 3 (\mod 5)$$n$$n$$n$$n$$a \in \mathbb Z$$n \in \mathbb N$$\overline a = \{b \in \mathbb Z \mid a \equiv b \ (\mod n)\} = \{a + kn \mi…</description>
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$\renewcommand{\mod}{\operatorname{mod}}$

Sometimes when you want to calculate something
you might only require the remainder of the answer when divided by some natural number $n$. With help of modular arithmetics finding the required remainder usually happens to be much easier than evaluating the original expression.$a, b$$n$$a \equiv b \ (\mod n)$$a$$b$$n$$a \mod n = b \mod n$$5638 \times 7891 + 2731 \times 3124$$17$$5638 \times 7891 + 2731 \times 3124 = 44489458 + 853164…</description>
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Let $f$ be a real-valued function defined on some subset of the real line $\mathbb R$. Informally speaking, a number $c$ is said to be the limit of $f$ in the process $x \to \infty$ if $f(x)$ comes arbitrarily close to $c$ as $x$ gets larger and larger. The precise definition of limit is as follows:$c \in \mathbb R$$\epsilon$$N$$x &gt; N$$|f(x)-c| &lt; \epsilon$$x$$\lim_{x \to \infty} f(x) = c$$c$$f$$x \to \infty$$M$$N$$x &gt; N$$f(x) &gt; M$$x \to \infty$$\lim_{x \to \infty} f(x) = \infty$$M$$N$$x &gt;…</description>
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