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(B) Let $S = \{1, 2, \dots, 25\}$. We want to find the number of pairs $(x, y)$ such that $x, y \in S$,$x 
eq y$,and $x + y \equiv 0 \pmod{5}$.First,

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$120$

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(B) Let $S = \{1, 2, \dots, 25\}$. We want to find the number of pairs $(x, y)$ such that $x, y \in S$,$x 
eq y$,and $x + y \equiv 0 \pmod{5}$.First,partition $S$ into subsets based on their remainder modulo $5$:$R_0 = \{5, 10, 15, 20, 25\}$ (size $5$)$R_1 = \{1, 6, 11, 16, 21\}$ (size $5$)$R_2 = \{2, 7, 12, 17, 22\}$ (size $5$)$R_3 = \{3, 8, 13, 18, 23\}$ (size $5$)$R_4 = \{4, 9, 14, 19, 24\}$ (size $5$)For $x+y$ to be divisible by $5$,the possible pairs of remainders $(r_x, r_y)$ are:$1$. $(0, 0)$: $x, y \in R_0$. Number of ways = $5 	imes 4 = 20$.$2$. $(1, 4)$: $x \in R_1, y \in R_4$. Number of ways = $5 	imes 5 = 25$.$3$. $(4, 1)$: $x \in R_4, y \in R_1$. Number of ways = $5 	imes 5 = 25$.$4$. $(2, 3)$: $x \in R_2, y \in R_3$. Number of ways = $5 	imes 5 = 25$.$5$. $(3, 2)$: $x \in R_3, y \in R_2$. Number of ways = $5 	imes 5 = 25$.Total ways = $20 + 25 + 25 + 25 + 25 = 120$.

Let $x$ and $y$ be distinct integers where $1 \leq x \leq 25$ and $1 \leq y \leq 25$. Then,the number of ways of choosing $x$ and $y$ such that $x + y$ is divisible by $5$ is $.........$.

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