Aptitude | Arithmetic Aptitude 2 | Question 2

There is bag of lots of integers from 1 to 10000, but you can only pick the numbers that are of 3-digit and also divisible by 3. What maximum sum of numbers you can pick.

(A)

165150

(B)

1600843

(C)

168132420

(D)

165322501

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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