$X=\{1,2,\dots,10\},x\rho y\Leftrightarrow x\equiv y(mod\hspace{0.2cm}3)$

Question

$X=\{1,2,\dots,10\},x\rho y\Leftrightarrow x\equiv y(mod\hspace{0.2cm}3)$

i.e $x,y$ have the same reminder when divided by $3$ ( it was actually written in the question).

I need to find the number of elements in $\rho$

Clearly $\rho$ is a reflexive relation,symmetric,transitive. I can see $|\rho|\ge 10$ due to reflexivity. Thank you for helping.

options were $40,36,34,33$

Answer 1

x=y  gives you 10 possibilities
   Since x≡y(mod3)
(x-y)/3=k  a constant.
So, x=3k+y
x=y+3 gives you 7 possibilities (using the numbers 1-10)
x=y+6 gives you 4 possibilities
x=y+9 gives you 1 possibilities
and since the relation is symmetrical
y=x+3 gives you 7 more possibilities
y=x+6 gives you 4 more
y=x+9 gives you 1 more.
This adds to 34. Those are my thoughts at this time. Hope it helps.

Answer 2

Formally, $$\rho\equiv\{(x,y)\in X\times X\,|\,x-y\text{ is divisible by 3}\}.$$

The elements of $\rho$ are: \begin{align*} \begin{array}{cccc} (1,1)&(1,4)&(1,7)&(1,10)\\ (2,2)&(2,5)&(2,8)&\\ (3,3)&(3,6)&(3,9)&\\ (4,1)&(4,4)&(4,7)&(4,10)\\ \vdots&\vdots&\vdots&\vdots\\ (10,1)&(10,4)&(10,7)&(10,10) \end{array}\end{align*} Enumerate all these elements. The answer is 34.