Let the relations $$R_1$$ and $$R_2$$ on the set $$X = \{1, 2, 3, \ldots, 20\}$$ be given by $$R_1 = \{(x, y) : 2x - 3y = 2\}$$ and $$R_2 = \{(x, y) : -5x + 4y = 0\}$$. If $$M$$ and $$N$$ be the minimum number of elements required to be added in $$R_1$$ and $$R_2$$, respectively, in order to make the relations symmetric, then $$M + N$$ equals

We need to find M + N where M and N are the minimum elements to add to make $$R_1$$ and $$R_2$$ symmetric.

To find the elements of $$R_1$$, note that $$R_1 = \{(x,y) : 2x - 3y = 2\}$$ where $$x, y \in \{1, 2, \ldots, 20\}$$. Solving: $$x = \frac{3y + 2}{2}$$. For $$x$$ to be a positive integer, $$3y + 2$$ must be even, so $$y$$ must be even.

Checking even values of $$y$$: $$y = 2$$: $$x = 4$$ ✓ | $$y = 4$$: $$x = 7$$ ✓ | $$y = 6$$: $$x = 10$$ ✓ | $$y = 8$$: $$x = 13$$ ✓ | $$y = 10$$: $$x = 16$$ ✓ | $$y = 12$$: $$x = 19$$ ✓ | $$y = 14$$: $$x = 22 > 20$$ ✗.

Thus $$R_1 = \{(4,2), (7,4), (10,6), (13,8), (16,10), (19,12)\}$$ — 6 pairs.

For symmetry in $$R_1$$, if $$(a,b) \in R_1$$ then $$(b,a)$$ must also be in $$R_1$$. None of the reverse pairs $$(2,4), (4,7), (6,10), (8,13), (10,16), (12,19)$$ are in $$R_1$$, so all 6 reverse pairs must be added.

M = 6

Next, to find the elements of $$R_2$$, observe that $$R_2 = \{(x,y) : -5x + 4y = 0\}$$, i.e., $$y = \frac{5x}{4}$$. For $$y$$ to be a positive integer, $$x$$ must be a multiple of 4.

Checking: $$x = 4$$: $$y = 5$$ ✓ | $$x = 8$$: $$y = 10$$ ✓ | $$x = 12$$: $$y = 15$$ ✓ | $$x = 16$$: $$y = 20$$ ✓ | $$x = 20$$: $$y = 25 > 20$$ ✗.

Hence $$R_2 = \{(4,5), (8,10), (12,15), (16,20)\}$$ — 4 pairs.

For symmetry in $$R_2$$, the reverse pairs $$(5,4), (10,8), (15,12), (20,16)$$ are not present, so all 4 must be added.

N = 4

Finally, $$M + N = 6 + 4 = 10$$. The correct answer is Option 4: 10.