2005 AMC 10B Problems/Problem 20
What is the average (mean) of all 5-digit numbers that can be formed by using each of the digits 1, 3, 5, 7, and 8 exactly once?
We first look at how many times each number will appear in each slot. If we fix a number in a slot, then there are ways to arrange the other numbers, so each number appears in each spot times. Therefore, the sum of all such numbers is Since there are such numbers, we divide to get
We can first solve for the mean for the digits and since each is away from each other. The mean of the numbers than can be solved using these digits is . The total amount of numbers that can be formed using these digits is . The sum of these numbers is . Now we can find out the total value that was gained by replacing the with a . We can start how be calculating the gain when the was in the ones digit. Since there are numbers with the in the ones digit and was gain from each of them, is the number gained. Then, we repeat this with the tens, hundreds, thousands, and ten thousands place, leading to a total of as the total amount that was gained. Subtract this amount from the sum of the digits using the instead of the to get . Finally, we divide this by to get the average.
The average value of the digits is . Values occur in every place so .
|2005 AMC 10B (Problems • Answer Key • Resources)|
|1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25|
|All AMC 10 Problems and Solutions|
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.