Let $$A = \{1, 2, 3, \ldots, 100\}$$. Let $$R$$ be a relation on $$A$$ defined by $$(x, y) \in R$$ if and only if $$2x = 3y$$. Let $$R_1$$ be a symmetric relation on $$A$$ such that $$R \subset R_1$$ and the number of elements in $$R_1$$ is $$n$$. Then the minimum value of $$n$$ is

We have the set $$A = \{1,2,3,\dots,100\}$$ and the relation $$R$$ defined by $$(x,y)\in R\iff 2x=3y\,. $$

First we rewrite the equation $$2x=3y$$ in parametric form. We note that $$\gcd(2,3)=1$$, so there exists an integer $$k$$ such that

$$x=3k,\quad y=2k\,. $$

Since $$x\in A$$, we have $$1\le 3k\le 100\,$$ which gives

$$1\le k\le \frac{100}{3}\quad\Longrightarrow\quad 1\le k\le 33\,. $$

Also, since $$y\in A$$, we have $$1\le 2k\le 100\,$$ which gives

$$1\le k\le 50\,. $$

Combining these two bounds on $$k$$ gives

$$1\le k\le 33\,, $$ so there are exactly $$33$$ integer values of $$k$$. Therefore, the number of ordered pairs in $$R$$ is $$33\,. $$

Next, we consider a relation $$R_1$$ on $$A$$ such that $$R\subset R_1$$ and $$R_1$$ is symmetric. By definition of symmetry:

$$\text{if }(x,y)\in R_1\text{ then }(y,x)\in R_1\,. $$

Since $$R\subset R_1$$, all $$33$$ pairs of the form $$(3k,2k)$$ must lie in $$R_1$$. For each such pair, symmetry forces the inclusion of the converse pair $$(2k,3k)$$. Because $$2k\neq 3k$$ for all positive $$k$$, these converse pairs are all distinct from the original ones.

Thus the minimal symmetric relation $$R_1$$ contains the original $$33$$ pairs plus the $$33$$ converse pairs, for a total of

$$n = 33 + 33 = 66\,. $$

Therefore, the minimum value of $$n$$ is 66.