im in year 7 so can someone help me with that modulo thing.. i nedd someone to explain it to me.

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- Aug 11th 2006, 01:58 AM #1

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- Aug 11th 2006, 06:17 AM #2

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Here's a brief Wikipedia explanation.

http://en.wikipedia.org/wiki/Modular_arithmetic

- Aug 11th 2006, 07:07 AM #3

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Sometimes for basic explanations Wikipedia is not ideal for it is complicated.

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$\displaystyle a,b,n$ are integers and $\displaystyle n\not = 0$

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When we write,

$\displaystyle a\equiv b (\mbox{mod } n)$

We mean that $\displaystyle a-b$ is divisible by $\displaystyle n$.

For example,

$\displaystyle 5\equiv 3 (\mbox{mod } 2)$ because, $\displaystyle 5-3=2$ which is divisible by two.

For example,

$\displaystyle 5\not \equiv 2 (\mbox{mod } 2)$ because $\displaystyle 5-2=3$ which is not divisible by 2.

- Aug 11th 2006, 07:19 AM #4

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You can think of doing modular artithmetic kind of the same way you would, say, figure time on a clock.

In modulo 12, 10+3 = 1

because 10+3=13, of course, but 13 mod 12 = 1 because 13 divided by 12 leaves a remainder of 1.

Similarly, on a clock, 10pm + 3 hours = 1 am, not 13pm.

And then 1:00 + 18 hours = 1:00 + 6 hours = 7:00.

Because 18 mod 12 = 6.

So you've been using modular arithmetic as long as you've been telling time without knowing it was modular arithmetic. That knowledge will hopefully make you more comfortable with using it now in other modulos than 12.

- Aug 11th 2006, 07:34 AM #5

- Aug 11th 2006, 07:47 AM #6

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- Aug 11th 2006, 07:55 AM #7
Year 7 means 11 years old if he is from England so would probably not be familiar with that notation. However it would not be part of the syllabus either so you never know.

If I'm right he need a different solution to his previous post

- Aug 11th 2006, 08:22 AM #8

- Aug 11th 2006, 08:27 AM #9

- Aug 11th 2006, 08:31 AM #10
It's just notation, so long as you can just solve problems and not mind!

(Actually, $\displaystyle a\equiv b \ {\rm mod}n$ means a and b are members of the "equivalence relation" defined by: "having the same residue when divided by n").

- Aug 11th 2006, 09:20 AM #11