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(B) We are given $S = \{(x, y) : x, y \in \mathbb{Z}, 0 \leq x, y \leq 18\}$.We need to find the number of pairs $(x, y) \in S$ such that $3x + 4y + 5

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$19$

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(B) We are given $S = \{(x, y) : x, y \in \mathbb{Z}, 0 \leq x, y \leq 18\}$.We need to find the number of pairs $(x, y) \in S$ such that $3x + 4y + 5 \equiv 0 \pmod{19}$.This is equivalent to $3x + 4y \equiv -5 \equiv 14 \pmod{19}$.Since $x, y \in \{0, 1, 2, \dots, 18\}$,for each $x \in \{0, 1, \dots, 18\}$,we have $4y \equiv 14 - 3x \pmod{19}$.Since $\gcd(4, 19) = 1$,for every $x$,there exists a unique $y \in \{0, 1, \dots, 18\}$ satisfying the congruence.Specifically,$y \equiv 4^{-1}(14 - 3x) \pmod{19}$.Since $4 	imes 5 = 20 \equiv 1 \pmod{19}$,the inverse of $4$ modulo $19$ is $5$.Thus,$y \equiv 5(14 - 3x) \equiv 70 - 15x \equiv 13 - 15x \equiv 13 + 4x \pmod{19}$.Since $0 \leq y \leq 18$,for each $x \in \{0, 1, \dots, 18\}$,there is exactly one $y$ such that $y = (13 + 4x) \pmod{19}$.Since there are $19$ possible values for $x$ (from $0$ to $18$),there are exactly $19$ such pairs $(x, y)$.

Let $S = \{(a, b) : a, b \in \mathbb{Z}, 0 \leq a, b \leq 18\}$. The number of elements $(x, y)$ in $S$ such that $3x + 4y + 5$ is divisible by $19$ is:

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