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
