The number of 3-digit numbers, that are divisible by either 2 or 3 but not divisible by 7 is

Q: Wait, why do we subtract $n(A \cap B)$ when calculating the union of A and B?

Great question! When we add the count of numbers divisible by 2 ($n(A)$) and the count of numbers divisible by 3 ($n(B)$), the numbers that are divisible by both 2 and 3 (i.e., multiples of 6) are counted twice—once in set A and once in set B. To correct this overcounting, we must subtract the intersection $n(A \cap B)$ exactly once.

Q: How do I know if I should use the formula $l = a + (n-1)d$?

Whenever you are counting numbers in a sequence with a constant difference (like multiples of 2, 3, or 7), you are dealing with an Arithmetic Progression (AP). The formula $l = a + (n-1)d$ is your most reliable tool to find the number of terms ($n$) when you know the first term ($a$), the last term ($l$), and the common difference ($d$).

Q: Why do we add back the triple intersection $n(A \cap B \cap C)$ in the final step?

This is the beauty of the Principle of Inclusion-Exclusion! When we calculate the 'bad' numbers (those divisible by 7 and either 2 or 3), we look at $(A \cap C) \cup (B \cap C)$. When we add $n(A \cap C)$ and $n(B \cap C)$, the numbers divisible by 2, 3, AND 7 (multiples of 42) are counted in both intersections. We added them twice, so we must subtract them once to get the correct count of 'bad' numbers.

Question

The number of 3-digit numbers, that are divisible by either 2 or 3 but not divisible by 7 is

Sigma Percentile · Accepted Answer

### The Art of Counting: Navigating Divisibility

Welcome, future engineer! Today, we are not just solving a problem; we are embarking on a journey of logical organization. Imagine you are standing in front of a massive library containing all 3-digit numbers, from 100 to 999.

Your task is to filter this library, keeping only the books that meet specific criteria. This is the essence of set theory and the Principle of Inclusion-Exclusion.

### Phase 1: Defining the Universe

First, let us define our universe. We are looking at 3-digit numbers, where the smallest is 100 and the largest is 999.

To find the total count, we use the simple formula:
$$Total = 999 - 100 + 1 = 900$$

This is our search space. We are looking for numbers divisible by 2 or 3, but we have a strict 'no-go' zone: numbers divisible by 7.

### Phase 2: The Power of Two and Three

Let $A$ be the set of numbers divisible by 2, and $B$ be the set of numbers divisible by 3. To find $n(A)$, we look at the sequence $100, 102, \dots, 998$.

Using the AP formula $l = a + (n-1)d$, we get:
$$998 = 100 + (n-1)2 \Rightarrow n(A) = 450$$

Similarly, for $B$, the sequence is $102, 105, \dots, 999$. Applying the same logic:
$$999 = 102 + (n-1)3 \Rightarrow n(B) = 300$$

If we simply add these, we double-count the multiples of 6. The intersection $n(A \cap B)$ consists of multiples of 6: $102, 108, \dots, 996$.

Solving the sequence:
$$996 = 102 + (n-1)6 \Rightarrow n(A \cap B) = 150$$

Thus, the number of elements in $A \cup B$ is:
$$n(A \cup B) = 450 + 300 - 150 = 600$$

### Phase 3: The Intruder

Now, we must address the constraint: no multiples of 7. Let $C$ be the set of numbers divisible by 7.

We need to remove any number from our set of 600 that is also in $C$. This means we need to find the size of $(A \cup B) \cap C$, which is equivalent to $(A \cap C) \cup (B \cap C)$.

First, let's find $n(A \cap C)$, the multiples of 14: $112, 126, \dots, 994$. 
$$994 = 112 + (n-1)14 \Rightarrow n(A \cap C) = 64$$

Next, $n(B \cap C)$, the multiples of 21: $105, 126, \dots, 987$. 
$$987 = 105 + (n-1)21 \Rightarrow n(B \cap C) = 43$$

Finally, the triple intersection $n(A \cap B \cap C)$, the multiples of 42: $126, 168, \dots, 966$. 
$$966 = 126 + (n-1)42 \Rightarrow n(A \cap B \cap C) = 21$$

### Phase 4: The Final Subtraction

Using the Principle of Inclusion-Exclusion for the 'bad' numbers:
$$n((A \cap C) \cup (B \cap C)) = 64 + 43 - 21 = 86$$

These 86 numbers are the intruders that must be removed from our initial 600. The final calculation is:
$$600 - 86 = 514$$

We have successfully filtered the library! The elegance of this method lies in its systematic approach. By breaking down complex constraints into manageable intersections, we turn a daunting problem into a clear, logical path.

Animated Solution for Mathematics - Sets and Relations: The number of 3-digit numbers, that are divisible by either 2 or 3 but not divisible by 7 is

Enter Numerical Value:

Visualized Solution

Understanding the Problem

Numbers Divisible by $2$ ( $n (A)$ )

Numbers Divisible by $3$ ( $n (B)$ )

Intersection of $2$ and $3$ ( $n (A \cap B)$ )

Union of $2$ or $3$ ( $n (A \cup B)$ )

The Constraint: Not Divisible by $7$

Divisible by $2$ and $7$ ( $n (A \cap C)$ )

Divisible by $3$ and $7$ ( $n (B \cap C)$ )

The Triple Intersection ( $n (A \cap B \cap C)$ )

Calculating the 'Bad' Numbers

Final Subtraction

Summary and Key Takeaway

The Sigma Insight: Types of Sets and Set Operations

The Art of Counting

Phase 1

Phase 2

Phase 3

Phase 4

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