Question

(a,b) is an integer solution for 45x-28y=130
How can we prove that a is an even number and (b) is a multiple of 5 ?

Answer 1

Look at the right hand side. It is equivalent to \(0 \pmod{5}\).

Answer 2

Thus, the left-hand side has to be \(0 \pmod{5}\) as well. Now, that means \(45x - 28y \equiv 0 \pmod 5\) implying that \(28y \equiv 0 \pmod{5}\). 28 itself is not divisible by 5, so it is obvious that \(y\) has to.

Answer 3

For the first part, use the mod 2.

Answer 4

and mod 2 for x, fairly enough thank you

Answer 5

You are welcome!

Answer 6

@ParthKohli could you please tell me how do i answer to the qs it says what are the reminders of n^2 divided by 4 ?

Answer 7

Ah. Do we have to prove the result as well? While we are on it, look at the remainders when n is even and when n is odd.

Answer 8

Yes 0 when it's even and 1 when odd

Answer 9

Yes, that's right :)

Answer 10

Actually i have big problem with understanding what Congruence is :)

Answer 11

Congruence in modular arithmetic? I had trouble in that once as well. Well, if you say that \(a \equiv 3 \pmod{8}\), it means that \(a\) will get the same remainder when divided by \(8\) as \(3\) does.

Answer 12

mmm r[3/8]=r[a/8]

Answer 13

\[19 \equiv 22 \pmod{3} \]The above is true because both of them get the same remainder when divided by \(3\).

Answer 14

Exactly. There are many solutions for that including \(3\).

Answer 15

I know all of the definitions of congruence but i still have problem using its rules

Answer 16

OK! First things first. Let me give you a pretty good example.

Why are ALL integers congruent to 0 OR 1 OR 2 OR 3 (mod 4)?

Answer 17

hh because 3,2,1 are smaller than 4 ?

Answer 18

Let's think about it this way:

ALL integers, when divided by 4, will get remainders just as 0 OR 1 OR 2 OR 3 get when divided by 4.

Answer 19

0 gets remainder 0 when divided by 4.
1 gets remainder 1.
2 gets remainder 2.
3 gets remainder 3.

So, we mean that the only remainders we can get are 0, 1, 2 or 3. That's true right?

Answer 20

i know that the reminder is always smaller than the divedent

Answer 21

So what I said is true.

Answer 22

Sure it's right

Answer 23

Now, what if I said that all the integers are always congruent to 5 OR 6 OR 7 OR 8?

Answer 24

mod what ?

Answer 25

mod 4.

Answer 26

Yes it's right, if you are talking about reminders make sure it's smaller than the dividend

Answer 27

As i know

Answer 28

Great job, what other properties?

Answer 29

what properties :) ?

Answer 30

ah greater than 0

Answer 31

As you told me, you had trouble understanding congruencies.

Answer 32

Yes

Answer 33

Where?

Answer 34

I didn't understand you which properties are you looking for ?

Answer 35

I meant rules. What rules?

Answer 36

ah you mean the rules that i didn't understand?

Answer 37

Yes, hehe

Answer 38

:) let me call it properties as you already said, i think that changing the number by its reminder is kinda, confusing me :)

Answer 39

Yup! Well, it's just a step higher than divisibility. You know how if two numbers are equal, they must be divisible by the same number? We're doing this a level higher and comparing the remainders. And we don't need to replace a number by its remainder in congruencies!

Answer 40

\[6 \equiv -1[7]\]
and
\[6\equiv-1[7]\]
so
\[-1\equiv-1[7]\]

Answer 41

Oh? Why are you confused in this?

Answer 42

Maybe you are trying to mean something like the following:\[1 \equiv 8 \pmod 7\]and\[8 \equiv 15 \pmod 7\]so\[1 \equiv 15 \pmod 7\]

Answer 43

I see this as i replaced the 6 by its reminder by 7, can you change this knowledge from my hand

Answer 44

You have to view a congruence relation a little differently. \(a \equiv b \pmod 7\) just means that they both get the same remainder. No replacements. :)

Answer 45

Yes Exactly its the base propertie i guess

Answer 46

Hmm... why does it not make sense to you? If a number gets the same remainder as the second number, and the second number gets the same as the third, then all the three get the same remainders.

Answer 47

Ok i think that i begin understanding the GOAL :)

Answer 48

if (a+b+c+d) sum of numbers has the same reminder as 6 when they are divided by 7
can i say that every one of a,b,c has the same reminder when divided by 7

Answer 49

as 6 sorry

Answer 50

Nope! You can't. Let me give you an example.

a = 1
b = 2
c = 3
d = 0

Answer 51

\[1+2+3+0\equiv6[7]\]

Answer 52

Yup.

Answer 53

But do all of them have the same remainder?

Answer 54

mm Nope

Answer 55

So the answer to your question is no.

Answer 56

what about the sum of tow

Answer 57

Nope, again! But you must be wondering how \(45x - 28y \equiv 0 \pmod{5}\) implies \(28y \equiv 0 \pmod{5}\)

Answer 58

mmm it's a multiple of numbers,right ?

Answer 59

because of 45x

Answer 60

Exactly! \(45x \equiv 0 \pmod{5}\)

Answer 61

Do you see why?

Answer 62

when you say that i view it as you replacing 0 by 45x

Answer 63

Yes, because both of them mean the same in mod 5.

Answer 64

But you don't necessarily have to replace it.

Answer 65

it's kinda rotation isn't it ?

Answer 66

Yup! That's the best way you can think of modular arithmetic.

Answer 67

numbers wrapping around it self

Answer 68

@ParthKohli I'm really begin understanding the concept :)

Answer 69

Yes, you've begun the journey to enlightenment!

Answer 70

@ParthKohli I don't know how to thank you enough, so thank you brother

Answer 71

I thoroughly enjoyed it! You're welcome, and thank YOU!

Answer 72

and sorry about the very bad english

Answer 73

Hehe, I'm not a native speaker either. No problem though! This is about math, not anything else. :)

Answer 74

@ganeshie8 you are enjoying the discussion i guess :)

Answer 75

Haha, he's a good friend.

Answer 76

yup very much :)

Answer 77

Ok guys thank you

Answer 78

I thought you had gone inactive! The fact that you stayed here and listened to us is so appreciable. :)

Answer 79

I'm trying to benefit 100% some people told me that i'm asking people to do my homeworks really my question is kinda homework but used it to start the discussion, i don't care to the solution of the equation because i studied another way to calculate the the solutions :)

Answer 80

Haha, you're a rare case on OS. :)

Answer 81

:)

Answer 82

I see, really some people they need to put their exercises books right here :)