Question

Hi folks, how can we solve for (a) this kind of Equations :
a= -1 mod 7
a= 0 mod 13

Answer 1

a = 0 mod 13

=>

a = 13k

Answer 2

good but why not 7k-1

Answer 3

plug that value in first equation

Answer 4

13k=-1 mod 7

Answer 5

13k = -1 mod 7

14k - k = -1 mod 7

-k = -1 mod 7

k = 1 mod 7

Answer 6

that gives k = 7m+1

Answer 7

plug this k value in the earlier 'a' value u got :-

a = 13k
= 13(7m+1)
= 13*7m + 13
= 13 mod (13*7)

Answer 8

thats the solution to given system of congruences,
see if that makes some sense..

Answer 9

How, you were looking for k and then a ??

Answer 10

in addition
how do you move from this :
14k - k = -1 mod 7
to this :
-k = -1 mod 7

Answer 11

ur ultimate goal is to find a

Answer 12

ok lets look at one question at a time :)

14k - k = -1 mod 7

Answer 13

yeah exactly so why you were looking for k

Answer 14

14k is divisible by 7,

so we say 14k = 0 mod 7

Answer 15

14k - k = -1 mod 7

0 - k = -1 mod 7

k = 1 mod 7

Answer 16

oh cool is it a rule ?

Answer 17

oly definition/rule we need to remember is this :

\(\large a \equiv b \mod n \)

means,

\(\large n | (a-b) \)

Answer 18

\(14k \equiv 0 \mod 7\)

so replace 14k with 0, cuz 14k leaves remainder 0 when divided by 7

Answer 19

for ur first question :-

a= 0 mod 13

=>

13 | (a-0)

13 | a

a = 13k

Answer 20

i mean changing 14k by 0 how do you define this

Answer 21

that means,

14k leaves a remainder 0 when divided by 7

Answer 22

14k = 0 = 28

in mod 7 world

Answer 23

mmm

Answer 24

so a = 91k+13

Answer 25

yes,
convince ur self by putting k = 0, 1, 2...

Answer 26

it should leave 0 when divided by 13, and -1 when divided by 7

Answer 27

yeah it's Working, in fact i can't even understand a lot of Congruent rules

Answer 28

takes some time to get use to notation,

but there are not many rules related to congruences.
reverything boils down to oly one definition :

a = b mod n
=>
n | (a-b)

Answer 29

always i say that the reminder of a/n is b

Answer 30

so i can replace the number by it's reminder ?

Answer 31

it's \[\equiv\]

Answer 32

yes cuz

\(a \equiv b \mod n\)

=>

\(n | (a-b) \)


that also gives us, \(n | (b-a)\)
or \(b \equiv a \mod n \)

so u can replace \(a\) and \(b\) in congruences

Answer 33

in simple words :-

if a-b is divisible by n,
then b-a is also divisible by n

so u can replace a and b

Answer 34

so:
if
a=3 mod 7
ab=5 mod 7
i can say that :
3a=5 mod 7
?

Answer 35

yes absolutely !
a = b mod n, and c = d mod n, then

ac = bd mod n

Answer 36

yeah but my case there is no (b times d)

Answer 37

GIven :-
a = 3 mod 7
ab = 5 mod 7

a = 7k + 3

(7k+3)b = 5 mod 7
3b = 5 mod 7

Answer 38

when u have, a = b mod n

u can replace \(a\) and \(b\) as u wish in other congruence expressions

Answer 39

Nice

Answer 40

but we didn't applied the rule that you have mantioned

Answer 41

and one thing that may help understand congruences faster :-

a = b mod n

dont think of it as : a/n leaves a remainder of b

Answer 42

correct way to think of it is :
(a-b) is divisible by n

Answer 43

this is exactly what i wast thinking of it as

Answer 44

or
a-b = nk

Answer 45

yeah i see that, that way of thinking will make it hard to understand congruences
(personal experience)

Answer 46

hhh, it's been while looking for somebody who wants to tell me this

Answer 47

and actually i begin understand whats happening right there

Answer 48

good :)

Answer 49

Thank you very much Brother, if you can find any think can help me please Contact me :)

Answer 50

np :) i see u doing great.. .just it requires more practice ...