被3整除的所有3位数字的总和是：
(A) 165150
(B) 164380
(C) 168420
(D) 165250答案： (A)
解释：
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. 这个问题的测验