How many numbers would reamin if all the numbers which are divisible by 3 are removed from 61 to 100 ?

Options

A. 13
B. 27
C. 26
D. 25

Explanation

To solve the problem of how many numbers remain if all the numbers divisible by 3 are removed from the range of 61 to 100, we can follow these steps:

### Step 1: Calculate the total numbers from 61 to 100
To find the total count of numbers from 61 to 100, we can use the formula for counting terms in a sequence:

\[ \text{Total Numbers} = \text{Last Term} - \text{First Term} + 1 \]

Here, the first term is 61 and the last term is 100.

\[ \text{Total Numbers} = 100 - 61 + 1 = 40 \]

### Step 2: Identify the numbers divisible by 3 in the range
Next, we need to find the numbers between 61 and 100 that are divisible by 3.

- The first number in this range that is divisible by 3 is 63.
- The last number in this range that is divisible by 3 is 99.

These numbers form an arithmetic progression (AP) where:
- First term (a) = 63
- Last term (l) = 99
- Common difference (d) = 3

To find the number of terms (n) in this AP, we can use the formula:

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

Substituting the values:

\[ n = \frac{99 - 63}{3} + 1 = \frac{36}{3} + 1 = 12 + 1 = 13 \]

### Step 3: Calculate the remaining numbers
Now, we subtract the count of numbers divisible by 3 from the total count of numbers:

\[ \text{Remaining Numbers} = \text{Total Numbers} - \text{Numbers Divisible by 3} \]

\[ \text{Remaining Numbers} = 40 - 13 = 27 \]

### Final Answer
Therefore, the number of numbers that remain after removing all the numbers divisible by 3 from 61 to 100 is **27**.

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