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(C) The total number of ways to select two numbers $x$ and $y$ from the set $\{1, 2, 3, \ldots, 10\}$ with replacement is $10 	imes 10 = 100$.We want

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$\frac{3}{10}$

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(C) The total number of ways to select two numbers $x$ and $y$ from the set $\{1, 2, 3, \ldots, 10\}$ with replacement is $10 	imes 10 = 100$.We want to find the number of pairs $(x, y)$ such that $|x^2 - y^2|$ is divisible by $6$.This condition is equivalent to $|(x - y)(x + y)|$ being divisible by $6$.We analyze the pairs $(x, y)$ by iterating through $x$ from $1$ to $10$ and finding $y$ such that $x^2 \equiv y^2 \pmod{6}$.The squares modulo $6$ are: $1^2 \equiv 1, 2^2 \equiv 4, 3^2 \equiv 3, 4^2 \equiv 4, 5^2 \equiv 1, 6^2 \equiv 0, 7^2 \equiv 1, 8^2 \equiv 4, 9^2 \equiv 3, 10^2 \equiv 4$.Grouping by values: $0: \{6\}$,$1: \{1, 5, 7\}$,$3: \{3, 9\}$,$4: \{2, 4, 8, 10\}$.Number of pairs $(x, y)$ such that $x^2 \equiv y^2 \pmod{6}$ is:For $x^2 \equiv 0$: $1^2 = 1$ pair $(6, 6)$.For $x^2 \equiv 1$: $3^2 = 9$ pairs (from $\{1, 5, 7\} 	imes \{1, 5, 7\}$).For $x^2 \equiv 3$: $2^2 = 4$ pairs (from $\{3, 9\} 	imes \{3, 9\}$).For $x^2 \equiv 4$: $4^2 = 16$ pairs (from $\{2, 4, 8, 10\} 	imes \{2, 4, 8, 10\}$).Total favorable outcomes $= 1 + 9 + 4 + 16 = 30$.Probability $= \frac{30}{100} = \frac{3}{10}$.

If two numbers $x$ and $y$ are chosen one after the other at random with replacement from the set of numbers $\{1, 2, 3, \ldots, 10\}$,then the probability that $|x^2 - y^2|$ is divisible by $6$ is

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