Number of 4-digit numbers that are less than or equal to 2800 and either divisible by 3 or by 11, is

Q: Why do we subtract the intersection $n(A \cap B)$? Isn't it just adding the two sets?

Great question! If you simply add $n(A)$ and $n(B)$, any number that is a multiple of both $3$ and $11$ (like $33, 66, 99, \dots$) gets counted once in $n(A)$ and again in $n(B)$. That means it is counted twice! To get the correct total, we must subtract that overlap once to correct the double-counting.

Q: How do I find the first and last terms of the AP for a given range?

For the first term $a$, find the smallest number $\ge 1000$ that is divisible by your divisor. For the last term $l$, find the largest number $\le 2800$ that is divisible by your divisor. You can use the remainder method: divide the boundary by the divisor, and adjust the number to make the remainder zero.

Q: Why is the intersection $A \cap B$ divisible by $33$?

Because $3$ and $11$ are coprime (they share no common factors other than $1$), any number divisible by both must be divisible by their product, which is $3 \times 11 = 33$. This is the definition of the Least Common Multiple (LCM) for coprime numbers.

Q: Is it possible to solve this without the AP formula?

You could technically list them, but that would take hours and lead to errors! The AP formula $n = \frac{l - a}{d} + 1$ is the standard, most efficient tool for counting multiples in a range. It is the backbone of this type of problem.

Question

Number of 4-digit numbers that are less than or equal to 2800 and either divisible by 3 or by 11, is equal to ____.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, my dear student. Today, we are not just solving a math problem; we are embarking on a journey to organize chaos. When you look at a problem asking for the number of integers divisible by $3$ or $11$ within a specific range, it is easy to feel overwhelmed.

But let us pause and breathe. Mathematics is the art of breaking down complexity into elegant, manageable pieces.

Let us define our playground. We are looking for four-digit numbers. The smallest four-digit number is $1000$. The problem gives us a strict upper bound: the numbers must be less than or equal to $2800$.

So, our effective range is the closed interval $[1000, 2800]$. This is our universe. Every number we consider must live within these walls.

### The Logic of Inclusion-Exclusion

Now, we need numbers divisible by $3$ OR $11$. In the language of set theory, this 'OR' is a union. Let $A$ be the set of numbers divisible by $3$, and $B$ be the set of numbers divisible by $11$.

We want to find the size of $A \cup B$. If we just add the number of multiples of $3$ and the number of multiples of $11$, we commit a classic error: double-counting.

Any number divisible by both $3$ and $11$—which means it is divisible by $33$—is counted in set $A$ and again in set $B$. To fix this, we use the **Principle of Inclusion-Exclusion**:

$$n(A \cup B) = n(A) + n(B) - n(A \cap B)$$

This formula is our compass.

### The Engine: Arithmetic Progressions

To find the number of elements in each set, we treat them as Arithmetic Progressions (AP). For any set of multiples of $k$ in a range $[a, l]$, the number of terms is given by:

$$n = \frac{l - a}{d} + 1$$

Here, $d$ is the common difference, which is simply $k$.

Let us find $n(A)$, the multiples of $3$. The first multiple of $3$ after $1000$ is $1002$. The largest multiple of $3$ before or at $2800$ is $2799$.

Using our formula:

$$n(3) = \frac{2799 - 1002}{3} + 1 = \frac{1797}{3} + 1 = 599 + 1 = 600$$

There are $600$ such numbers.

Now, for $n(B)$, the multiples of $11$. The first multiple of $11$ after $1000$ is $1001$. The largest multiple of $11$ before or at $2800$ is $2794$.

Applying the formula:

$$n(11) = \frac{2794 - 1001}{11} + 1 = \frac{1793}{11} + 1 = 163 + 1 = 164$$

We have $164$ multiples of $11$.

### The Overlap: The Final Piece

Finally, we must calculate the intersection $A \cap B$, which are the multiples of $33$. The first multiple of $33$ after $1000$ is $1023$. The largest multiple of $33$ before or at $2800$ is $2772$.

Our calculation:

$$n(33) = \frac{2772 - 1023}{33} + 1 = \frac{1749}{33} + 1 = 53 + 1 = 54$$

We have $54$ numbers that were counted twice.

### The Grand Finale

We have all our components. We take the count of $A$, add the count of $B$, and subtract the overlap:

$$600 + 164 - 54 = 764 - 54 = 710$$

There you have it! Through logical deduction and the power of arithmetic progressions, we have arrived at the answer: **$710$**.

Never fear the complexity of a problem; simply peel back the layers, one by one, and the solution will reveal itself.

Animated Solution for Mathematics - Sequence and Series: Number of 4-digit numbers that are less than or equal to 2800 and either divisible by 3 or by 11, is equal to ____.

Enter Numerical Value:

Visualized Solution

Defining the Range

Strategy: Inclusion-Exclusion

The Overlap Concept

Numbers Divisible by 3

Calculating $n (3)$

Numbers Divisible by 11

Calculating $n (11)$

The Overlap: Divisible by 33

Calculating $n (33)$

Final Calculation

The Sigma Insight: Arithmetic Progression (A.P.)

Analyzing the Setup

The Logic of Inclusion-Exclusion

The Engine

The Overlap

The Grand Finale

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