Let's re-share it!

We know that when a number is divided by $3$ it can leave remainders: $0,1.2$ $Case 1:x\equiv0(mod3),y\equiv0(mod3)\Longrightarrow\\x^{2}\equiv0(mod3),y^{2}\equiv0(mod3).$ $\Longrightarrow\ x^{2}-y^{2}\equiv0(mod3)$ $Case2:x\equiv1(mod3),y\equiv1(mod3)\\\Longrightarrow\ x^{2}\equiv1(mod3),y^{2}\equiv1(mod3)\Longrightarrow\ x^{2}-y^{2}\equiv0(mod3)$ $Case3:x\equiv2(mod3),y\equiv2(mod3$ \ $\Longrightarrow x^{2}\equiv1(mod3),y^{2}\equiv1(mod3)\Longrightarrow\ x^{2}-y^{2}\equiv0(mod3).$ Things were pretty straightforward till now,the next case is a little different. $Case4:x\equiv1(mod3),y\equiv2(mod3)\Longrightarrow\\\ x^{2}\equiv1(mod3),y^{2}\equiv1(mod3)\\\Longrightarrow\ x^{2}-y^{2}\equiv0(mod3)$ Number of ways of case1:since there are a 100 numbers which are divisible by 3 in the range 1 to 300.Selecting two numbers can be done by $\dbinom{100}{2}.$ The same thing is done for cases 2 and 3.Now,in case 4 we have to choose one number from the group of numbers that are $\equiv1(mod3)$ and one number from the group of numbers that are $\equiv2(mod3)$ .This can be done in $\dbinom{100}{1}*\dbinom{100}{1}$ .The total number of ways is $\dbinom{300}{2}.$ Just calculate and you get the answer.

Please reply to me about the problem of this question.

Lemma: If $3 \nmid x,y$ , then $3 \mid x^{2} - y^{2}$ .

Proof: Let $x = 3k \pm 1, y = 3m \mp 1, \forall k,m \in \mathbb{Z}$

$\therefore x^{2}-y^{2} = (x-y)(x+y) = (3k\pm 1 + 3p\mp 1)(3k\pm 1 - (3p\mp 1)) = 3(k+m)(3k+3p\pm 2)$ such that $(k+m)(3k+3p\pm 2) \in \mathbb{Z}$ .

Therefore, $3 \nmid x,y$ $\rightarrow$ $3 \mid x^{2} - y^{2}$ .

Case1: $3 \mid x,y$

• Choose $2$ from $100$ for $x,y$ , $\dbinom{100}{2}$

Because it's stated that $x > y$ , choosing $2$ of them doesn't require permutations because one of them is obviously less than the other.

Case2: $3 \nmid x,y$ from lemma

• Choose $2$ from $200$ for $x,y$ , $\dbinom{200}{2}$

Total number of ways $= \dbinom{100}{2} + \dbinom{200}{2}$ .

Sample spaces

• Choose $2$ from $300$ for $x,y$ , $\dbinom{300}{2}$

Possible outcomes $= \dbinom{300}{2}$

Therefore, probability $\displaystyle = \frac{\dbinom{100}{2} + \dbinom{200}{2}}{\dbinom{300}{2}} = \frac{497}{897}$ . ~~~