Let A = {1, 2, 3, ldots, 100}. Let R be a rel | vedclass

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(B) The relation $R$ is defined as $(x, y) \in R \iff 2x = 3y$,which implies $y = \frac{2}{3}x$. Since $x, y \in \{1, 2, \ldots, 100\}$,$x$ must be a

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$66$

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(B) The relation $R$ is defined as $(x, y) \in R \iff 2x = 3y$,which implies $y = \frac{2}{3}x$. Since $x, y \in \{1, 2, \ldots, 100\}$,$x$ must be a multiple of $3$. Let $x = 3k$,then $y = 2k$. For $1 \le 2k \le 100$,we have $k \le 50$. For $1 \le 3k \le 100$,we have $k \le 33$. Thus,$k$ can take values from $1$ to $33$. The elements of $R$ are $\{(3, 2), (6, 4), (9, 6), \ldots, (99, 66)\}$. The number of elements in $R$ is $n(R) = 33$. For $R_1$ to be a symmetric relation such that $R \subset R_1$,for every $(x, y) \in R$,the pair $(y, x)$ must also be in $R_1$. Since $R$ is not symmetric (e.g.,$(3, 2) \in R$ but $(2, 3) 
otin R$),we must include all $(y, x)$ pairs in $R_1$. Thus,$R_1 = R \cup R^{-1} = \{(3, 2), (6, 4), \ldots, (99, 66), (2, 3), (4, 6), \ldots, (66, 99)\}$. The number of elements in $R_1$ is $n = n(R) + n(R^{-1}) = 33 + 33 = 66$.

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:

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