Understanding the Sum of Integers from 1 to 100 Divisible by 2 or 5

In mathematics, the sum of integers from 1 to 100 that are divisible by 2 or 5 is a common problem. This article explores the methodologies to solve such problems, discussing various techniques and formulas used to determine the sum of specific integer sets. By understanding the principles behind these sums, you can apply similar techniques to solve a wide range of mathematical problems.

Introduction to the Problem

The integers from 1 to 100 that are divisible by 2 or 5 can be represented as a set of numbers. These numbers form a specific pattern that can be exploited to find the sum efficiently. The sum of these numbers is a classic problem that involves both arithmetic and combinatorial techniques.

Identification of Specific Numbers

The integers from 1 to 100 that are divisible by 2 or 5 are as follows:

Numbers divisible by 2: 2, 4, 6, 8, ..., 100 Numbers divisible by 5: 5, 10, 15, 20, ..., 100 Numbers divisible by both 2 and 5 (i.e., 10): 10, 20, 30, 40, ..., 100

Technically, for a number to be divisible by 2 or 5, we include all numbers in the sets above, but we must subtract the numbers that are divisible by both to avoid double-counting.

Mathematical Formulas and Techniques

We can break the problem into two parts: the sum of numbers divisible by 2 and the sum of numbers divisible by 5. Then, we subtract the sum of numbers divisible by 10 (since they are counted twice).

Sum of Numbers Divisible by 2 or 5

Let's find the sum of the numbers divisible by 2:

S_2 2 4 6 ... 100

This is an arithmetic series with the first term (a 2) and the common difference (d 2). The number of terms is 50. The sum (S_2) can be calculated using the formula for the sum of an arithmetic series:

S_n n/2 (2a (n-1)d)

Plugging in the values:

S_2 50/2 (2 times; 2 (50-1) times; 2) 25 times; (4 98) 2550

Similarly, for the numbers divisible by 5:

S_5 5 10 15 ... 100

This is also an arithmetic series with the first term (a 5) and the common difference (d 5). The number of terms is 20. Using the same formula:

S_5 20/2 (2 times; 5 (20-1) times; 5) 10 times; (10 95) 1050

For the numbers divisible by both 2 and 5 (i.e., 10):

S_{10} 10 20 30 ... 100

This is again an arithmetic series with the first term (a 10) and the common difference (d 10). The number of terms is 10. Using the same formula:

S_{10} 10/2 (2 times; 10 (10-1) times; 10) 5 times; (20 90) 550

Now, to find the sum of the numbers divisible by 2 or 5, we subtract (S_{10}) from the sum of (S_2) and (S_5):

S S_2 S_5 - S_{10} 2550 1050 - 550 3050

Generalized Technique

A more generalized approach involves summing the series directly without breaking it down. We can use the principle of inclusion-exclusion:

S sum_{j1}^{50} 2j sum_{j1}^{33} 3j - sum_{j1}^{16} 6j

This formula avoids double-counting numbers that are divisible by 6. Let's break it down:

S 2(1 2 3 ... 50) 3(1 2 3 ... 33) - 6(1 2 3 ... 16)

Using the arithmetic series sum formula:

sum_{j1}^{n} j n(n 1)/2

Plugging in the values:

S 2(50 times; 51/2) 3(33 times; 34/2) - 6(16 times; 17/2) 2550 1683 - 816 3417

However, this approach slightly deviates from the simplified formula. The simplified approach yields:

S 3050

Python Program Implementation

A simple Python program to find the sum using a loop:

sum  0
for number in range(1, 101):
    if number % 2  0 or number % 5  0:
        sum   number
print('The sum is {}'.format(sum))

The output of this program is:

The sum is 3050

Conclusion

By employing the principles of arithmetic series and the inclusion-exclusion principle, we can efficiently find the sum of integers from 1 to 100 that are divisible by 2 or 5. Understanding these techniques not only helps in solving this particular problem but also equips you with tools that can be applied to a wide range of mathematical problems.