Let A = {n ∈ N : n is a 3-digit number}, B = {9k + 2 : k ∈ N} and C = {9k + ℓ : k ∈ N} for some ℓ (0

Q: Wait, why can we just add the sums of $A \cap B$ and $A \cap C$?

Great question! We can add them because the sets $B$ and $C$ are disjoint. Since $B = \{9k + 2\}$ and $C = \{9k + \ell\}$ with $0 < \ell < 9$ and $\ell \neq 2$, no number can belong to both sets. If a number $n$ were in both, it would have to satisfy $n \equiv 2 \pmod 9$ and $n \equiv \ell \pmod 9$, which is impossible if $\ell \neq 2$. Thus, the intersection of the two sets is empty, and the sum of their union is the sum of their individual parts.

Q: How do I know there are exactly 100 terms in the intersection?

To find the number of terms, look at the inequality $100 \le 9k + r \le 999$. Subtract $r$ from all sides: $100 - r \le 9k \le 999 - r$. Divide by $9$: $\frac{100-r}{9} \le k \le \frac{999-r}{9}$. Since $k$ must be an integer, you count the number of possible integer values for $k$. For $r=2$, $10.88 \le k \le 110.77$, so $k$ ranges from $11$ to $110$. The number of terms is $110 - 11 + 1 = 100$. This logic holds for any $r$ between $1$ and $8$.

Q: Why is the first term $99 + \ell$ and not just $\ell$?

Remember, our universe is Set $A$, which contains only 3-digit numbers ($100 \le n \le 999$). If we just used $\ell$, we might get a 1-digit or 2-digit number. We need the smallest number of the form $9k + \ell$ that is greater than or equal to $100$. Since $99$ is the largest 2-digit multiple of $9$, the next number of the form $9k + \ell$ is $99 + \ell$. This ensures we stay within the 3-digit range.

Question

Let $A = \{n \in \mathbb{N} : n 	ext{ is a 3-digit number}\}$, $B = \{9k + 2 : k \in \mathbb{N}\}$ and $C = \{9k + \ell : k \in \mathbb{N}\}$ for some $\ell (0 < \ell < 9)$. If the sum of all the elements of the set $A \cap (B \cup C)$ is $274 	imes 400$, then $\ell$ is equal to

Sigma Percentile · Accepted Answer

### Analyzing the Universe of Numbers

Imagine you are standing on a vast, infinite number line. Your task is to zoom in on a specific, finite segment: the three-digit numbers. This is our Set $A$, our universe for this problem.

It starts at $100$ and ends at $999$. The total count of numbers is calculated as:
$$999 - 100 + 1 = 900$$
We have $900$ integers to play with. Now, we introduce two subsets, $B$ and $C$, defined by their remainders when divided by $9$.

### The Red Dots: Analyzing Set $B$

Set $B$ consists of numbers of the form $9k + 2$. We are interested in the intersection $A \cap B$, which means we only care about the numbers in this form that fall within our $900$-number universe.

To find the first term, we test values of $k$. If $k = 11$, we get $9(11) + 2 = 101$. This is our first red dot.

To find the last term, we look for the largest $k$ such that $9k + 2 \le 999$. That gives us $k = 110$, resulting in:
$$9(110) + 2 = 992$$

We have an arithmetic progression: $101, 110, 119, \dots, 992$. Using the formula for the number of terms $n$:
$$n = \frac{992 - 101}{9} + 1 = 100$$
We have $100$ red dots. The sum of these, $S_1$, is:
$$S_1 = \frac{100}{2}(101 + 992) = 50 	imes 1093 = 54650$$

### The Blue Dots: The Disjoint Union

Now, consider Set $C$, defined as $9k + \ell$. Because $B$ and $C$ have different remainders modulo $9$, they are disjoint—they share no common elements.

This is a powerful realization! It means the sum of the union $A \cap (B \cup C)$ is simply the sum of the individual intersections:
$$S_{total} = S_1 + S_2$$

We are given $S_{total} = 274 	imes 400 = 109600$. With $S_1 = 54650$, we can easily find $S_2$:
$$S_2 = 109600 - 54650 = 54950$$
This $S_2$ is the sum of all elements in $A \cap C$.

### The Final Reveal: Solving for $\ell$

We know $A \cap C$ is also an arithmetic progression with $100$ terms. The first term is $99 + \ell$ and the last term is $990 + \ell$.

Applying the sum formula again:
$$S_2 = \frac{100}{2} [(99 + \ell) + (990 + \ell)] = 54950$$

Simplifying this, we get:
$$50 [1089 + 2\ell] = 54950$$
Dividing by $50$, we find:
$$1089 + 2\ell = 1099$$

Subtracting $1089$ from both sides, we get $2\ell = 10$, which leads us to the elegant conclusion:
**$\ell = 5$**

You have successfully navigated the arithmetic progression, understood the disjoint nature of the sets, and solved for the unknown. This is the power of systematic thinking in JEE mathematics.

Animated Solution for Mathematics - Sequence and Series: Let $A = {n \in N : n is a 3-digit number}$ , $B = {9 k + 2 : k \in N}$ and $C = {9 k + ℓ : k \in N}$ for some $ℓ (0 < ℓ < 9)$ . If the sum of all the elements of the set $A \cap (B \cup C)$ is $274 \times 400$ , then $ℓ$ is equal to

Enter Numerical Value:

Visualized Solution

Defining Set $A$ : The Universe

Analyzing Set $B$ : Remainder $2$

Finding the Last Term of $A \cap B$

Counting Terms in $A \cap B$

Calculating Sum $S_{1}$

Total Sum Calculation

Finding Sum $S_{2}$

Defining Terms of $A \cap C$

Setting up the Equation for $ℓ$

Solving for $ℓ$

Final Conclusion

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

Analyzing the Universe of Numbers

The Red Dots

The Blue Dots

The Final Reveal

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Enter Numerical Value:

Visualized Solution

Defining Set .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A: The Universe

Analyzing Set .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B: Remainder .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2

Finding the Last Term of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A∩B

Counting Terms in .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A∩B

Calculating Sum .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S1​