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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.

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>a \pmod{m} \cdot b \pmod{m} \equiv (a \cdot b) \pmod{m} </math>
+==Useful Theorems==
+Fermat's Little Theorem:For a prime <math>p</math> and a number <math>a</math> such that <math>a\ne{p}</math>, <math>a^{p-1}\equiv 1 \pmod{p}</math>.
+Wilson's Theorem: For a prime <math>p</math>, <math> (p-1)! \equiv -1 \pmod p</math>.
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