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Permutations and Combinations: Mathematics | JEE Main

Q: Why do we subtract floor(999/k) instead of floor(1000/k)?

To count the number of integers in divisible by , we take the number of multiples up to , which is , and subtract the number of multiples up to , which is . Since 4-digit numbers start at 1000, , so .

Number of 4-digit numbers that are less than or equal to

2800

and either divisible by

3

or by

11

, is equal to

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Step-by-step solutionView

Correct answer

The number of 4-digit numbers less than or equal to 2800 that are divisible by 3 or 11 is 710.

Option analysis

Why each option works or fails

Reviewed route

Solution

StepWorking

01given

We need to find the number of 4-digit integers $x$ such that $1000 \le x \le 2800$ and $x$ is divisible by $3$ or by $11$ .

02goal

Find $n(A \cup B)$ , where $A$ is the set of multiples of $3$ in $[1000, 2800]$ and $B$ is the set of multiples of $11$ in $[1000, 2800]$ .

03approach

By PIE, $n(A \cup B) = n(A) + n(B) - n(A \cap B)$ , where $A \cap B$ is the set of multiples of $\operatorname{lcm}(3, 11) = 33$ . Note that $2800$ is not divisible by $3$ or $11$ , so the interval of 4-digit numbers can be taken as $[1000, 2800]$ or $[1000, 2799]$ . For any integer $k$ , the number of multiples in $[1000, 2800]$ is $\lfloor 2800/k \rfloor - \lfloor 999/k \rfloor$ .

04execute

Compute $n(A)$ , the number of multiples of 3 in $[1000, 2800]$ : $n(A) = \lfloor 2800/3 \rfloor - \lfloor 999/3 \rfloor = 933 - 333 = 600$ .

05execute

Compute $n(B)$ , the number of multiples of 11 in $[1000, 2800]$ : $n(B) = \lfloor 2800/11 \rfloor - \lfloor 999/11 \rfloor = 254 - 90 = 164$ .

06execute

Compute $n(A \cap B)$ , the number of multiples of 33 in $[1000, 2800]$ : $n(A \cap B) = \lfloor 2800/33 \rfloor - \lfloor 999/33 \rfloor = 84 - 30 = 54$ .

07execute

Apply the Principle of Inclusion-Exclusion: $n(A \cup B) = n(A) + n(B) - n(A \cap B) = 600 + 164 - 54 = 710$ .

✓verify

Check boundary points: $2800 = 3 \times 933 + 1 = 11 \times 254 + 6$ , neither multiple, so using 2800 or 2799 yields identical counts: $\lfloor 2799/3 \rfloor - \lfloor 999/3 \rfloor = 933 - 333 = 600$ ; $\lfloor 2799/11 \rfloor - \lfloor 999/11 \rfloor = 254 - 90 = 164$ ; $\lfloor 2799/33 \rfloor - \lfloor 999/33 \rfloor = 84 - 30 = 54$ . $600 + 164 - 54 = 710$ . Correct.

Hints that build this answer step by step

What range of integers corresponds to 4-digit numbers less than or equal to 2800?
$1000 \le n \le 2800$
Which formula gives the number of integers in $[1000, 2800]$ divisible by 3 or 11 using the Principle of Inclusion-Exclusion?
$n(3) + n(11) - n(33)$
How many integers in the range $[1000, 2800]$ are multiples of 3, 11, and 33 respectively?
$n(3) = 600$ , $n(11) = 164$ , $n(33) = 54$
What is the final count $n(3) + n(11) - n(33)$ ?
710

✓ Source and academic review

Question type: Numerical
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 9 September 2026

Quick checks

Students also ask

Why do we subtract floor(999/k) instead of floor(1000/k)?

To count the number of integers in $[a, b]$ divisible by $k$ , we take the number of multiples up to $b$ , which is $\lfloor b/k \rfloor$ , and subtract the number of multiples up to $a-1$ , which is $\lfloor (a-1)/k \rfloor$ . Since 4-digit numbers start at 1000, $a=1000$ , so $a-1 = 999$ .

Answer

The number of 4-digit numbers less than or equal to 2800 that are divisible by 3 or 11 is 710.

Why each option works or fails

Step-by-step solution

given: We need to find the number of 4-digit integers $x$ such that $1000 \le x \le 2800$ and $x$ is divisible by $3$ or by $11$ .
goal: Find $n(A \cup B)$ , where $A$ is the set of multiples of $3$ in $[1000, 2800]$ and $B$ is the set of multiples of $11$ in $[1000, 2800]$ .
approach: By PIE, $n(A \cup B) = n(A) + n(B) - n(A \cap B)$ , where $A \cap B$ is the set of multiples of $\operatorname{lcm}(3, 11) = 33$ . Note that $2800$ is not divisible by $3$ or $11$ , so the interval of 4-digit numbers can be taken as $[1000, 2800]$ or $[1000, 2799]$ . For any integer $k$ , the number of multiples in $[1000, 2800]$ is $\lfloor 2800/k \rfloor - \lfloor 999/k \rfloor$ .
execute: Compute $n(A)$ , the number of multiples of 3 in $[1000, 2800]$ : $n(A) = \lfloor 2800/3 \rfloor - \lfloor 999/3 \rfloor = 933 - 333 = 600$ .
execute: Compute $n(B)$ , the number of multiples of 11 in $[1000, 2800]$ : $n(B) = \lfloor 2800/11 \rfloor - \lfloor 999/11 \rfloor = 254 - 90 = 164$ .
execute: Compute $n(A \cap B)$ , the number of multiples of 33 in $[1000, 2800]$ : $n(A \cap B) = \lfloor 2800/33 \rfloor - \lfloor 999/33 \rfloor = 84 - 30 = 54$ .
execute: Apply the Principle of Inclusion-Exclusion: $n(A \cup B) = n(A) + n(B) - n(A \cap B) = 600 + 164 - 54 = 710$ .
verify: Check boundary points: $2800 = 3 \times 933 + 1 = 11 \times 254 + 6$ , neither multiple, so using 2800 or 2799 yields identical counts: $\lfloor 2799/3 \rfloor - \lfloor 999/3 \rfloor = 933 - 333 = 600$ ; $\lfloor 2799/11 \rfloor - \lfloor 999/11 \rfloor = 254 - 90 = 164$ ; $\lfloor 2799/33 \rfloor - \lfloor 999/33 \rfloor = 84 - 30 = 54$ . $600 + 164 - 54 = 710$ . Correct.

Permutations and Combinations: Mathematics | JEE Main

What feels right?

The number of 4-digit numbers less than or equal to 2800 that are divisible by 3 or 11 is 710.

Why each option works or fails

Solution

01given

02goal

03approach

04execute

05execute

06execute

07execute

✓verify

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