Aptitude | Arithmetic Aptitude 2 | Question 2

The sum of all 3 digit numbers divisible by 3 is:

**(A)** 165150

**(B)** 164380

**(C)** 168420

**(D)** 165250

**Answer:** **(A)** **Explanation:**

All 3 digit numbers divisible by 3 are : 102, 105, 108, 111, ..., 999. This is an A.P. with first element 'a' as 102 and difference 'd' as 3. Let it contains n terms. Then, 102 + (n - 1) x 3 = 999 102 + 3n-3 = 999 3n = 900 or n = 300 Sum of AP = n/2 [2*a + (n-1)*d] Required sum = 300/2[2*102 + 299*3] = 165150.