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