Question

Find the sum of integers from 1 to 100 which are divisible by 2 or 3.

• A. 3412
• B. 3417
• C. 3422
• D. 3401

Answer 1

The correct option is B 3417

The numbers which are divisible by 2 from 1 to 100 are
2, 4, ... 98, 100.

The numbers which are divisible by 3 from 1 to 100 are
3, 6, ... 96, 99.

In the above series, the numbers that are multiple of both 2 and 3 (i.e. 6), occur twice.
Therefore,
Sum of integers which are divisible by 2 or 3 from 1 to 100
= (Sum of integers divisible by 2 from 1 to 100) + (Sum of integers divisible by 3 from 1 to 100) − (Sum of integers divisible by 6 from 1 to 100)

= (2 + 4 + .... + 100) + (3 + 6 + .... + 99) − (6 + 12 + ... 96)
= 2(1 + 2 + .... + 50) + 3(1 + 2 + ... +33) − 6(1 + 2 + .... + 16)
= $2\times \left(\frac{50\times 51}{2}\right)$ + $3\times \left(\frac{33\times 34}{2}\right)$ - $6\times \left(\frac{16\times 17}{2}\right)$
[$\because$ Sum of first n natural numbers = $\frac{n(n+1)}{n}$]
= 2550 + 1683 − 816
= 4233 − 816
= 3417