The sum of the integers from 1 to `100` which are not divisible by 3 or `5` is

Options

A. 2489
B. 4735
C. 2632
D. 2317

Answer

To find the sum of integers from 1 to 100 that are not divisible by 3 or 5, we will follow these steps:

### Step 1: Calculate the sum of the first 100 natural numbers.
The formula for the sum of the first n natural numbers is given by:
\[ S_n = \frac{n(n + 1)}{2} \]
For \( n = 100 \):
\[ S_{100} = \frac{100 \times 101}{2} = 5050 \]

### Step 2: Calculate the sum of multiples of 3 up to 100.
The multiples of 3 up to 100 are 3, 6, 9, ..., 99. This is an arithmetic progression (AP) where:
- First term \( a = 3 \)
- Common difference \( d = 3 \)
- Last term \( l = 99 \)

To find the number of terms \( n \):
\[ l = a + (n - 1)d \implies 99 = 3 + (n - 1) \cdot 3 \]
Solving for \( n \):
\[ 99 - 3 = (n - 1) \cdot 3 \implies 96 = (n - 1) \cdot 3 \implies n - 1 = 32 \implies n = 33 \]

Now, we can calculate the sum of these multiples:
\[ S_n = \frac{n}{2} \cdot (a + l) = \frac{33}{2} \cdot (3 + 99) = \frac{33}{2} \cdot 102 = 33 \cdot 51 = 1683 \]

### Step 3: Calculate the sum of multiples of 5 up to 100.
The multiples of 5 up to 100 are 5, 10, 15, ..., 100. This is also an AP where:
- First term \( a = 5 \)
- Common difference \( d = 5 \)
- Last term \( l = 100 \)

To find the number of terms \( n \):
\[ l = a + (n - 1)d \implies 100 = 5 + (n - 1) \cdot 5 \]
Solving for \( n \):
\[ 100 - 5 = (n - 1) \cdot 5 \implies 95 = (n - 1) \cdot 5 \implies n - 1 = 19 \implies n = 20 \]

Now, we can calculate the sum of these multiples:
\[ S_n = \frac{n}{2} \cdot (a + l) = \frac{20}{2} \cdot (5 + 100) = 10 \cdot 105 = 1050 \]

### Step 4: Calculate the sum of multiples of both 3 and 5 (i.e., multiples of 15) up to 100.
The multiples of 15 up to 100 are 15, 30, 45, ..., 90. This is an AP where:
- First term \( a = 15 \)
- Common difference \( d = 15 \)
- Last term \( l = 90 \)

To find the number of terms \( n \):
\[ l = a + (n - 1)d \implies 90 = 15 + (n - 1) \cdot 15 \]
Solving for \( n \):
\[ 90 - 15 = (n - 1) \cdot 15 \implies 75 = (n - 1) \cdot 15 \implies n - 1 = 5 \implies n = 6 \]

Now, we can calculate the sum of these multiples:
\[ S_n = \frac{n}{2} \cdot (a + l) = \frac{6}{2} \cdot (15 + 90) = 3 \cdot 105 = 315 \]

### Step 5: Calculate the final sum \( S \).
Using the principle of inclusion-exclusion:
\[ S = S_{100} - S_{3} - S_{5} + S_{15} \]
Substituting the values we calculated:
\[ S = 5050 - 1683 - 1050 + 315 = 5050 - 2733 + 315 = 2632 \]

### Final Answer:
The sum of integers from 1 to 100 that are not divisible by 3 or 5 is **2632**.

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