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(A) We need to find the number of integers $x$ such that $1000 \le x \le 2800$ and $x$ is divisible by $3$ or $11$.Let $A$ be the set of numbers divis

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$710$

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(A) We need to find the number of integers $x$ such that $1000 \le x \le 2800$ and $x$ is divisible by $3$ or $11$.Let $A$ be the set of numbers divisible by $3$ and $B$ be the set of numbers divisible by $11$ in the range $[1000, 2800]$.The number of elements is given by $|A \cup B| = |A| + |B| - |A \cap B|$.For set $A$ (divisible by $3$): The smallest number is $1002$ and the largest is $2799$. Using $a_n = a + (n-1)d$,we have $2799 = 1002 + (n-1)3$,so $1797 = 3(n-1)$,$n-1 = 599$,$n = 600$.For set $B$ (divisible by $11$): The number of multiples of $11$ up to $2799$ is $\lfloor \frac{2799}{11} \rfloor = 254$. The number of multiples of $11$ up to $999$ is $\lfloor \frac{999}{11} \rfloor = 90$. Thus,$|B| = 254 - 90 = 164$.For set $A \cap B$ (divisible by $33$): The number of multiples of $33$ up to $2799$ is $\lfloor \frac{2799}{33} \rfloor = 84$. The number of multiples of $33$ up to $999$ is $\lfloor \frac{999}{33} \rfloor = 30$. Thus,$|A \cap B| = 84 - 30 = 54$.Finally,$|A \cup B| = 600 + 164 - 54 = 710$.

The number of $4$-digit numbers that are less than or equal to $2800$ and are either divisible by $3$ or by $11$ is equal to $............$.

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