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(B) Let $A$ be the set of numbers divisible by $3$ in the range $\{1, 2, \ldots, 2000\}$.$n(A) = \lfloor \frac{2000}{3} \rfloor = 666$.Let $B$ be the

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$214$

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(B) Let $A$ be the set of numbers divisible by $3$ in the range $\{1, 2, \ldots, 2000\}$.$n(A) = \lfloor \frac{2000}{3} floor = 666$.Let $B$ be the set of numbers divisible by $7$ in the range $\{1, 2, \ldots, 2000\}$.$n(B) = \lfloor \frac{2000}{7} floor = 285$.Let $A \cap B$ be the set of numbers divisible by both $3$ and $7$,i.e.,divisible by $	ext{lcm}(3, 7) = 21$.$n(A \cap B) = \lfloor \frac{2000}{21} floor = 95$.Using the principle of inclusion-exclusion,$n(A \cup B) = n(A) + n(B) - n(A \cap B) = 666 + 285 - 95 = 856$.The probability $p = \frac{n(A \cup B)}{2000} = \frac{856}{2000}$.Therefore,$500p = 500 	imes \frac{856}{2000} = \frac{856}{4} = 214$.

$A$ number is chosen at random from the set $\{1, 2, 3, \ldots, 2000\}$. Let $p$ be the probability that the chosen number is a multiple of $3$ or a multiple of $7$. Then the value of $500p$ is . . . . . .

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