[SOLVED] Z18 Calculation

**ronaldo_07** · March 2nd 2009, 06:26 PM

$\frac{2}{7}+\frac{7}{13}=\frac{[2]X[13]+[7]X[7]}{[7]X[13]}$

Im not sure how to multiply this out please show steps on how to do this. Thanks

**TheEmptySet** · March 2nd 2009, 06:56 PM

Originally Posted by ronaldo_07

$\frac{2}{7}+\frac{7}{13}=\frac{[2]X[13]+[7]X[7]}{[7]X[13]}$

Im not sure how to multiply this out please show steps on how to do this. Thanks

From the Euclidean algorithm (or trial and error)

$7^{-1}=13 \mod(18)$ and

$13^{-1}=7 \mod(18)$

so we end up with
$13 \cdot 2 +7\cdot 7 =26+49=75 \equiv 3 \mod(18)$

**ronaldo_07** · March 3rd 2009, 03:22 AM

How did you get from 75 to 3mod(17)?

**rajr** · March 3rd 2009, 07:01 AM

Hi as you mentioned it's not 3 mod 17, actually it is 3 mod 18.

take 75 and when you try to divide 75 by 18, you will get 3 as a remainder and hence 3 mod 18. is it helpful...?

**ThePerfectHacker** · March 3rd 2009, 09:32 AM

Here is another way. To find inverse of 7 you need to solve $7x\equiv 1(\bmod 18)$ .
This is equivalent to $7x\equiv 1 - 2\cdot 18(\bmod 18)\implies 7x\equiv -35(\bmod 18) \implies$ $x\equiv -5(\bmod 18)\implies x\equiv 13(\bmod 18)$ .

Thread: [SOLVED] Z18 Calculation

LinkBack

Thread Tools

Display

[SOLVED] Z18 Calculation

Similar Math Help Forum Discussions

Help with calculation...

GDP calculation

SST Calculation

[SOLVED] Compound Interest Calculation

pKa and pKb calculation

Search Tags