JEE Mathematics 2023 Question with Solution

Given: We need the number of $4$-digit numbers from $1000$ to $2799$ that are divisible by $3$ or $11$.

Find: The total count using the inclusion-exclusion principle.

Numbers divisible by $3$ form an arithmetic progression from $1002$ to $2799$:

$$1002 + (n - 1) \times 3 = 2799$$

So,

$$(n - 1) \times 3 = 1797$$ $$n - 1 = 599$$ $$n = 600$$

Hence, $n(3) = 600$.

For numbers divisible by $11$ between $1000$ and $2799$:

$$\left\lfloor \frac{2799}{11} \right\rfloor = 254$$ $$\left\lfloor \frac{999}{11} \right\rfloor = 90$$

Therefore,

$$n(11) = 254 - 90 = 164$$

Numbers divisible by both $3$ and $11$ are divisible by $33$:

$$\left\lfloor \frac{2799}{33} \right\rfloor = 84$$ $$\left\lfloor \frac{999}{33} \right\rfloor = 30$$

Therefore,

$$n(33) = 84 - 30 = 54$$

Now apply inclusion-exclusion:

$$n(3) + n(11) - n(33)$$ $$600 + 164 - 54 = 710$$

Therefore, the required number is $710$.

Given: Count the $4$-digit numbers less than or equal to $2800$ that are divisible by $3$ or $11$.

Find: Total count.

Step 1: Find the number of $4$-digit numbers divisible by $3$. The range is from $1000$ to $2799$. The first such number is $1002$ and the last is $2799$.

$$1002 + (n - 1) \times 3 = 2799$$ $$n = 600$$

So, there are $600$ numbers divisible by $3$.

Step 2: Find the number divisible by $11$.

$$\left\lfloor \frac{2799}{11} \right\rfloor = 254$$ $$\left\lfloor \frac{999}{11} \right\rfloor = 90$$ $$254 - 90 = 164$$

So, there are $164$ numbers divisible by $11$.

Step 3: Find the number divisible by both, that is, divisible by $33$.

$$\left\lfloor \frac{2799}{33} \right\rfloor = 84$$ $$\left\lfloor \frac{999}{33} \right\rfloor = 30$$ $$84 - 30 = 54$$

So, there are $54$ numbers divisible by $33$.

Step 4: Use inclusion-exclusion.

$$n(3) + n(11) - n(33) = 600 + 164 - 54 = 710$$

Therefore, the answer is $710$.