There are 27 different three-digit integers that can be form

Hi All,

We're told that there are 27 different three-digit integers that can be formed using only the digits 1, 2 and 3 and that all 27 of the integers were listed. We're asked for the SUM of those 27 different integers. This question has a great series of built-in 'logic shortcuts' that don't require a lot of advanced math (although a bit of arithmetic will still be required to get to the correct answer).

To start, we know that the smallest number would be 111 and the largest would be 333. Without too much effort, you should be able to figure out that...

9 of the numbers begin with a '1'
9 of the numbers begin with a '2'
9 of the numbers begin with a '3'

So when we add up JUST the 100s, we have.... 9(100) + 9(200) + 9(300) = 5400

Next, consider ALL of the 10s that we have... 27 of them (and some are 20s and some are 30s). That would add HUNDREDS (well over 270+) to the total sum. Based on how the answer choices are written, there's only one answer that makes sense...

Final Answer: E

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

BTGmoderatorDC wrote:There are 27 different three-digit integers that can be formed using only the digits 1, 2 and 3. If all 27 of the integers were listed, what would their sum be?

A. 2,704
B. 2,990
C. 5,404
D. 5,444
E. 5,994

Each of the digits 1, 2 and 3 appears 9 times in each of the hundreds, tens and units places. For example, the digit 1 appears 9 times in the hundreds place since there are 9 numbers in the 100s. Therefore, the sum of these 27 numbers must be:

100 x 9 + 200 x 9 + 300 x 9 + 10 x 9 + 20 x 9 + 30 x 9 + 1 x 9 + 2 x 9 + 3 x 9

(100 + 200 + 300) x 9 + (10 + 20 + 30) x 9 + (1 + 2 + 3) x 9

600 x 9 + 60 x 9 + 6 x 9

(600 + 60 + 6) x 9

666 x 9

5,994

Alternate Solution:

Since there are 27 3-digit numbers, we know that 9 of them will have a “3” in the hundreds place (e.g., 312, 321, 331, 313, etc.), and 9 of them will have a “2” in the hundreds place (e.g., 222, 231, 213, etc.), and 9 of them will have a “1” in the hundreds place (e.g., 111, 123, 121, etc.). Thus, if we add ONLY the hundreds places, we see that we will have a sum of 9 x 300 + 9 x 200 + 9 x 100 = 2700 + 1800 + 900 = 5400. We easily can eliminate choices A, B, and C. We can also eliminate choice D because we are assured that, once we add up the 27 numbers in the tens position and the 27 numbers in the units position, we will far exceed the answer in choice D. The only choice that remains is choice E.

Answer: E