vinay1983 wrote:What is the sum of all 3 digit positive integers that can be formed using the digits 1, 5, and 8, if the digits are allowed to repeat within a number?

A. 126
B. 1386
C. 3108
D. 308
E. 13986

Here's a different approach:

First, as Sul noted, there are 27 possible 3-digit numbers that we can create.
We know this because we have 3 options (1, 5 or 8) for the hundreds digit, 3 options for the tens digit, and 3 options for the units digit.
So, by the Fundamental Counting Principle (FCP), the number of ways to create a 3-digit number = (3)(3)(3) = 27

Aside: For more information about the FCP, we have a free video on the subject: https://www.gmatprepnow.com/module/gmat-counting?id=775

IMPORTANT: Of the 27 different 3-digit numbers, 9 of them will have 1 in the hundreds place, 9 of them will have 5 in the hundreds place, and 9 of them will have 8 in the hundreds place.

Now we'll use a bit of number sense. We have 27 3-digit numbers and THE AVERAGE value of these 27 numbers is WELL OVER 200. So, the sum of the 27 numbers must be GREATER THAN (27)(200)
In other words, the sum of the 27 numbers is GREATER THAN 5400

So, the answer must be E

Cheers,
Brent

Hi vinay1983,

Brent makes some very good points in his explanation, but you can take it a step further (and do even less math) to answer this question.

Since you have 9 numbers that start with 8, we would be dealing with....

8-- x 9 = 7200+

Just by thinking about those 9 numbers, you know the total is going to be over 7200, so the answer has to be E. There's no need to do any other steps.

GMAT assassins aren't born, they're made,
Rich