Difference between revisions of "User:Temperal/The Problem Solver's Resource6"

Revision as of 19:49, 4 October 2007

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

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

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

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

@@ Line 10: / Line 10: @@
 *<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==
+<!-- to be filled in soon -->
 ==Properties==
 For any number there will be only one congruent number modulo <math>m</math> between <math>0</math> and <math>m-1</math>.
@@ Line 17: / Line 18: @@
 <math>a \pmod{m} + b \pmod{m} \equiv (a + b) \pmod{m}</math>
-<math>a \pmod{m} - b \pmod{m} \equiv (a - b) \pmod{m} </math>
+<math>a \pmod{m} - b \pmod{m} \equiv (a - b) \pmod{m}</math>
 <math>a \pmod{m} \cdot b \pmod{m} \equiv (a \cdot b) \pmod{m} </math>