Key Concepts and Formulas

• Principle of Inclusion-Exclusion: For two sets A and B, the sum of elements in A or B is given by $S(A \cup B) = S(A) + S(B) - S(A \cap B)$.
• Arithmetic Progression (AP): A sequence with a constant difference between consecutive terms.
• Sum of an AP: $S_n = \frac{n}{2}(a + l)$, where $n$ is the number of terms, $a$ is the first term, and $l$ is the last term.
• Number of terms in an AP: $n = \frac{l - a}{d} + 1$, where $d$ is the common difference.

Step-by-Step Solution

Step 1: Calculate the sum of integers from 1 to 100 that are divisible by 2 ($S_2$).

• We need to find the sum of the arithmetic progression: 2, 4, 6, ..., 100.
• Here, the first term $a = 2$, the last term $l = 100$, and the common difference $d = 2$.
• The number of terms is $n = \frac{100 - 2}{2} + 1 = \frac{98}{2} + 1 = 49 + 1 = 50$.
• Therefore, the sum is $S_2 = \frac{50}{2}(2 + 100) = 25 \times 102 = 2550$.
• This represents the sum of all multiples of 2 between 1 and 100.

Step 2: Calculate the sum of integers from 1 to 100 that are divisible by 5 ($S_5$).

• We need to find the sum of the arithmetic progression: 5, 10, 15, ..., 100.
• Here, the first term $a = 5$, the last term $l = 100$, and the common difference $d = 5$.
• The number of terms is $n = \frac{100 - 5}{5} + 1 = \frac{95}{5} + 1 = 19 + 1 = 20$.
• Therefore, the sum is $S_5 = \frac{20}{2}(5 + 100) = 10 \times 105 = 1050$.
• This represents the sum of all multiples of 5 between 1 and 100.

Step 3: Calculate the sum of integers from 1 to 100 that are divisible by both 2 and 5 (i.e., divisible by 10) ($S_{10}$).

• We need to find the sum of the arithmetic progression: 10, 20, 30, ..., 100. Since numbers divisible by both 2 and 5 are divisible by their least common multiple (LCM), which is 10, we are summing the multiples of 10.
• Here, the first term $a = 10$, the last term $l = 100$, and the common difference $d = 10$.
• The number of terms is $n = \frac{100 - 10}{10} + 1 = \frac{90}{10} + 1 = 9 + 1 = 10$.
• Therefore, the sum is $S_{10} = \frac{10}{2}(10 + 100) = 5 \times 110 = 550$.
• This represents the sum of all multiples of 10 between 1 and 100. We need to subtract this to avoid double-counting.

Step 4: Apply the Principle of Inclusion-Exclusion to find the total sum.

• Using the formula $S_{\text{total}} = S_2 + S_5 - S_{10}$, we have:
• $S_{\text{total}} = 2550 + 1050 - 550 = 3600 - 550 = 3050$.
• This final calculation combines the individual sums and then adjusts for the numbers that were counted twice, providing the accurate sum of all integers from 1 to 100 that are divisible by 2 or 5.

Common Mistakes & Tips

• Remember to use the Principle of Inclusion-Exclusion when dealing with "or" conditions to avoid double-counting.
• Don't forget the "+1" when calculating the number of terms in an arithmetic progression.
• The LCM of two numbers is crucial when considering numbers divisible by both.

Summary

We used the Principle of Inclusion-Exclusion to calculate the sum of integers from 1 to 100 that are divisible by 2 or 5. We found the sums of multiples of 2, multiples of 5, and multiples of 10, and then applied the formula $S_{\text{total}} = S_2 + S_5 - S_{10}$ to get the final answer.

The final answer is $\boxed{3050}$, which corresponds to option (B).