Question

If 2 numbers A and B are selected from the set of first 100 natural numbers then in how many ways,

Answer 1

\(a^2-b^2\) is divisible by 3?

Answer 2

Lets try..
Numbers are of form: 3k,3k+1,3k+2
Squared numbers are of form:
\[(3k)^2=9k^2\]
\[(3k+1)^2=9k^2+6k+1=3k(3k+2)+1\]
\[(3k+2)^2=9k^2+12k+4=3k(3k+4)+4\]

Answer 3

Remember, \(r\in \{0,1,2\}\)

Answer 4

yup

Answer 5

4 = 3+1

Answer 6

what do you mean?

Answer 7

\[
4\equiv 1 \pmod 3
\]

Answer 8

didn't get what that meant since morning :p

Answer 9

You don't know what modulus is, but you know what divisibility is?!

Answer 10

I know modulus but what are you relating here I'm not getting

Answer 11

I'm saying that the remainder of \((3k+2)^2\) is \(1\) rather than \(4\).

Answer 12

how ? :/

Answer 13

okay yes

Answer 14

got it
4=1+3 :P

Answer 15

You're killing me!

Answer 16

So squared numbers are of form :
3k,3k+1,3k+1 (do i need to write this twice?)

Answer 17

Writing it twice can help.

Answer 18

Let us consider all the cases now :D
1)A of form 3k,B of form 3k
2)Both of form 3k+2
3)A of form 3k+1,B of 3k+2
4)A of form 3k+2,B of form 3k+1

Answer 19

No, which combinations of these remainders has remainder 0?

Answer 20

Now^

Answer 21

Did I leave any case?

Answer 22

There were \(3\times 3\) cases.

Answer 23

i am still getting confused :/

Answer 24

There are \(9\) cases.

Answer 25

The question is "a^2-b^2" divisible by 3
where did we use that info?I mean same solution for a^2+b^2 ?
why did we compute the squared form of numbers? we dont have 3k+2 in that but then how can we assume it in the cases

Answer 26

\((3k+2)^2=3k'+1\)

Answer 27

Okay, so when considering the TOTAL cases, what we consider is how many remainders \(a\) could have (3) and how many \(b\) can have(3).
Combine this and you get \(3\times3=9\) cases.

Answer 28

The next step is to find out which of those \(9\) cases end up being divisible by three.

Answer 29

yes

Answer 30

As it turns out, there are only two possible remainders you can get after squaring: 0 and 1.

Answer 31

yes!

Answer 32

Now, we need to consider the cases:
0+0 = 0
0+1 = 1
1+0 = 1
1+1 = 2

Answer 33

Now, 1+1=2 isn't divisible by 3 because it doesn't have remainder of 0.

But hypothetically if it WAS divisible by 3..
Then it since there are 2 ways to have remainder 1 ( 3k+1, 3k+2 )

Then the ways to get 1+1 is \(2\times2=4\)

Answer 34

Fortunately the only case divisible by \(3\) is:
\(0+0=1\)
There is only \(1\times 1=1\) ways for this to happen.

Answer 35

thought:
will we get same answer if question says a^2 "+" b^2 ?

Answer 36

Oh wait.. it should be -?

Answer 37

Yeah, that is a big difference!

Answer 38

0-0 = 0
1-0 = 1
0-1 = -1
1-1 = 0

Answer 39

So 0-0 and 1-1 are divisible by 3 in this case.

Answer 40

0-0 happens \(1\times1=1\) ways
1-1 happens \(2\times2=4\) ways

Answer 41

So \(5\) out of \(9\) ways!

Answer 42

that is what I was thinking :| but
\[(3k+1)^2-(3k+2)^2=1+6k-(4+6k)=-3\]
if we take A of form 3k+1 and B of form 3k+2

Answer 43

-3 :|

Answer 44

\(\color{blue}{\text{Originally Posted by}}\) @DLS
that is what I was thinking :| but
\[(3k+1)^2-(3k+2)^2=1+6k-(4+6k)=-3\]
if we take A of form 3k+1 and B of form 3k+2
\(\color{blue}{\text{End of Quote}}\)
\[
(3k+1)^2-(3k+2)^2=3k'+1 - (3k'+1)=0
\]

Answer 45

what just happened?

Answer 46

Ugh...

Answer 47

Okay, so get this: we ONLY need to worry about the REMAINDER

Answer 48

yes true so :o

Answer 49

So why are you worrying about other pellet?

Answer 50

About other \(\text{pellet}\)?

Answer 51

About other \(\text{sh}
\text{it}\)

Answer 52

but it is giving negative remainder .-.

Answer 53

If it is giving a negative remainder then: 3-r = ?

Answer 54

3-1 = 2, the remainder is 2

Answer 55

Or rather remainder -1 becomes remainder 2

Answer 56

okay i got evt

Answer 57

What I would do is this: only consider the cases where a>b

Answer 58

then double that result for b>a

Answer 59

seems a good idea