The ones that have a remainder of 2 will all be 1 less than the multiples of 3. (Example 102-1 = 101 , 999-1 = 998)

So we have an arithmetic progression: first term = 101, last term = 998. Difference between 2 terms = 3

a(n) = a(1) + (n-1) * d
=> 998 = 101 + (n-1) * 3
=> n = 300

Sn = n/2(a1 + an) = 300/2 * (101+998) = 150 * 1099 ... Ans

This is just an arithmetic sequence.

first such number is 101. last number is 998.
101, 104,................995,998

sum = n/2(2a+(n-1)*d)
n will be same as number of multiples of 3 between 100 and 1000.

for 1000 there are 333 and for 100 there are 33. so n for us is 300

sum = 300/2(202+(299)*3) = 164850

Please Let me know OA

What is the sum of all 3 digit numbers that leave a remainder of '2' when divided by 3?

is there technique to resolve quickly please?

soln: basically we want number of the form 3k+2

1st 3 digit number will be 100/3=33 so for k=33 you will get 101 as first number now it will be AP series with 3 as common difference3
last number will be 999/3=333 for k =333 you will get 4 digit number i.e 333*3 + 2=1001
so last 3 digit number will be for k=332 and number will be 333*3+2=998

so all together k will take 332-33+1=300 values
so sum [= n/2*(a+l) ap formula
150(101+998)=150(1099)=29850


in short number will be of form 3k+2
first number=101 last number 998
then arithmetic progression sum formula

typo error answer will be 164850

thks guys

have you maths background?

not exactly but i like numbers
specially number properties

gmat is nightmare whether you are not good in numbers or fast !

I did my ba in economics in 2001, so long time ago o:)