The number of 3 digit numbers, that are divisible by either 3 or 4 but not divisible by 48, is (1) 4

Q: Wait, why is the total count $900$ and not $899$?

That is a classic trap! When counting integers in a range $[a, b]$, the number of elements is $b - a + 1$. For the range $[100, 999]$, it is $999 - 100 + 1 = 900$. Always remember to include both endpoints!

Q: Why do we use the Principle of Inclusion-Exclusion here?

Great question! If we just added the count of numbers divisible by $3$ and the count of numbers divisible by $4$, we would count numbers divisible by both (like $12, 24, 36...$) twice. The Principle of Inclusion-Exclusion, $n(3 \cup 4) = n(3) + n(4) - n(3 \cap 4)$, corrects this by subtracting the intersection once.

Q: Why is the LCM of $3$ and $4$ equal to $12$?

Since $3$ and $4$ are co-prime (their greatest common divisor is $1$), their least common multiple is simply their product, $3 \times 4 = 12$. Any number divisible by both must be a multiple of $12$.

Q: Why do we subtract the count of $48$ from the union?

The problem asks for numbers divisible by $3$ or $4$ *but not* $48$. Since $48$ is a multiple of $12$, every number divisible by $48$ is already inside the set of numbers divisible by $12$, which is the intersection of our $3$ and $4$ sets. Thus, all multiples of $48$ are already included in our union of $450$. To satisfy the 'not divisible by $48$' condition, we must remove them.

Question

The number of 3 digit numbers, that are divisible by either 3 or 4 but not divisible by 48, is

Options:

(A) 472

(B) 432

(C) 507

(D) 400

Sigma Percentile · Accepted Answer

### Analyzing the Universe of Numbers

Imagine you are standing on the vast number line, looking at the stretch from $100$ to $999$. This is our universe, our set $U$.

To find how many numbers live here, we do not just subtract $100$ from $999$; we must be precise. The count is $999 - 100 + 1 = 900$. This is the foundation of our journey.

### The Blue and Green Circles: Divisibility

Now, let us consider the numbers divisible by $3$. In any sequence of consecutive integers, every third number is a multiple of $3$.

So, out of our $900$ numbers, the count is:

$$n(3) = \frac{900}{3} = 300$$

Let us visualize this as a blue circle. Next, we look for numbers divisible by $4$. Similarly, every fourth number is a multiple of $4$. The count is:

$$n(4) = \frac{900}{4} = 225$$

This is our green circle.

### The Overlap: The Power of LCM

Here is where it gets interesting. Some numbers are divisible by both $3$ and $4$. These numbers live in the overlap of our blue and green circles.

A number divisible by both $3$ and $4$ must be divisible by their least common multiple, $	ext{LCM}(3, 4) = 12$. So, we need to count the multiples of $12$ in our range:

$$n(12) = \frac{900}{12} = 75$$

These $75$ numbers have been counted twice—once in the blue circle and once in the green circle.

### The Union: Inclusion-Exclusion

To find the total number of elements divisible by either $3$ or $4$, we use the Principle of Inclusion-Exclusion. We add the counts of the two sets and subtract the intersection to fix the double-counting:

$$n(3 \cup 4) = n(3) + n(4) - n(3 \cap 4) = 300 + 225 - 75 = 450$$

We now have $450$ numbers that satisfy the first part of our condition.

### The Red Constraint: The $48$ Trap

But wait, the problem has a final, sneaky constraint: 'not divisible by $48$'. Where do these numbers live?

Since $48$ is a multiple of $12$, every number divisible by $48$ is automatically divisible by $12$. This means the set of multiples of $48$ is entirely contained within the intersection we just analyzed.

We need to find how many such numbers exist in our range $[100, 999]$. The multiples of $48$ are $48 	imes 3 = 144$, $48 	imes 4 = 192$, and so on, up to $48 	imes 20 = 960$.

The number of terms is $20 - 3 + 1 = 18$.

### The Final Victory

We have $450$ numbers in our union, but $18$ of them are multiples of $48$, which we must exclude.

The final calculation is simple but satisfying:

$$450 - 18 = 432$$

We have navigated the sets, accounted for the overlap, and successfully dodged the trap. The answer is **$432$**.

Animated Solution for Mathematics - Sets and Relations: The number of 3 digit numbers, that are divisible by either 3 or 4 but not divisible by 48, is

Select Answer:

Visualized Solution

Universal Set $U$

Divisibility by $3$

Divisibility by $4$

Intersection: Divisible by $12$

Union: Divisible by $3$ or $4$

The Exception: Divisible by $48$

Counting $n (48)$

Final Answer

The Sigma Insight: Types of Sets and Set Operations

Analyzing the Universe of Numbers

The Blue and Green Circles

The Overlap

The Union

The Red Constraint

The Final Victory

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