Question

The number of natural numbers divisible by 3 between 1 to 100 is:

• A. 11
• B. 33
• C. 36
• D. 22

Hint:

For any range starting from 1 to $N$, the count of numbers divisible by $k$ is simply the integer division of $N$ by $k$, i.e., $\lfloor N/k \rfloor$. Here, $100 / 3 = 33$ remainder 1, yielding 33 directly.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
To find the count of natural numbers divisible by a given divisor within a specific range, we can utilize the properties of an Arithmetic Progression (AP).
The natural numbers divisible by $3$ between $1$ and $100$ form a sequential progression starting from the smallest multiple of 3 and ending at the largest multiple of 3 below 100.
Key Formula or Approach:
The general term ($a_n$) of an Arithmetic Progression is defined as: \[ a_n = a + (n - 1)d \] where:
- $a$ is the first term.
- $d$ is the common difference.
- $n$ is the total number of terms.
- $a_n$ is the last term of the progression.

Step 2: Detailed Explanation:
Identify the components of our sequence:
The first multiple of 3 greater than 1 is: \[ a = 3 \] The common difference is: \[ d = 3 \] The largest multiple of 3 less than 100 is: \[ a_n = 99 \] Substitute these values into the AP formula to solve for $n$: \[ 99 = 3 + (n - 1) \cdot 3 \] Subtract 3 from both sides: \[ 96 = (n - 1) \cdot 3 \] Divide by 3: \[ n - 1 = \frac{96}{3} \] \[ n - 1 = 32 \] Add 1 to both sides: \[ n = 33 \] Alternatively, this can be solved using the greatest integer function (floor function): \[ n = \left\lfloor \frac{100}{3} \right\rfloor = \lfloor 33.33 \rfloor = 33 \]

Step 3: Final Answer:
The total number of natural numbers divisible by 3 between 1 and 100 is 33.