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The Problem Solver's Resource

Introduction \| Other Tips and Tricks \| Methods of Proof \| You are currently viewing page 6.

$n\equiv a\pmod{b}$ if $n$ is the remainder when $a$ is divided by $b$ to give an integral amount.

Occasionally, if two equivalent expressions are both modulated by the same number, the entire equation will be followed by the modulo.

For any number there will be only one congruent number modulo $m$ between $0$ and $m-1$ .

If $a\equiv b \pmod{m}$ and $c \equiv d \pmod{m}$ , then $(a+c) \equiv (b+d) \pmod {m}$ .

$a \pmod{m} + b \pmod{m} \equiv (a + b) \pmod{m}$

$a \pmod{m} - b \pmod{m} \equiv (a - b) \pmod{m}$

$a \pmod{m} \cdot b \pmod{m} \equiv (a \cdot b) \pmod{m}$

Fermat's Little Theorem:For a prime $p$ and a number $a$ such that $a\ne{p}$ , $a^{p-1}\equiv 1 \pmod{p}$ .

Wilson's Theorem: For a prime $p$ , $(p-1)! \equiv -1 \pmod p$ .

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 *<math>n\equiv a\pmod{b}</math> if <math>n</math> is the remainder when <math>a</math> is divided by <math>b</math> to give an integral amount.
 ==Special Notation==
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+Occasionally, if two equivalent expressions are both modulated by the same number, the entire equation will be followed by the modulo.
 ==Properties==
 For any number there will be only one congruent number modulo <math>m</math> between <math>0</math> and <math>m-1</math>.