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. 


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