JEE 2023 — Mathematics PYQ

Solution

To find the count of such numbers, we need to consider the range of 4-digit numbers starting from 1000 up to 2800. We use the Principle of Inclusion-Exclusion:

1. Numbers divisible by 3 ($n(A)$)

The first 4-digit number divisible by 3 is $1002$, and the last one $\le 2800$ is $2799$.

Using the Arithmetic Progression (AP) formula $a_n=a+(n-1)d$:

• $a=1002,l=2799,d=3$

• $2799=1002+(n_1-1)3$

• $1797=(n_1-1)3$

• $n_1-1=599⟹n_1=600$

  $n(A)=600$

2. Numbers divisible by 11 ($n(B)$)

The first 4-digit number divisible by 11 is $1001$, and the last one $\le 2800$ is $2794$.

• $a=1001,l=2794,d=11$

• $2794=1001+(n_2-1)11$

• $1793=(n_2-1)11$

• $n_2-1=163⟹n_2=164$

  $n(B)=164$

3. Numbers divisible by both 3 and 11 ($n(A\cap B)$)

Numbers divisible by both 3 and 11 must be divisible by their LCM, which is 33.

The first such number is $1023$, and the last one $\le 2800$ is $2772$.

• $a=1023,l=2772,d=33$

• $2772=1023+(n_3-1)33$

• $1749=(n_3-1)33$

• $n_3-1=53⟹n_3=54$

  $n(A\cap B)=54$

Final Calculation

Now, applying the inclusion-exclusion principle:

Answer: The number of such 4-digit numbers is 710.