Difference between revisions of "2005 AMC 10B Problems/Problem 20"

Problem

What is the average (mean) of all 5-digit numbers that can be formed by using each of the digits 1, 3, 5, 7, 8 and exactly once?

$\mathrm{(A)}48000\qquad\mathrm{(B)}49999.5\qquad\mathrm{(C)}53332.8\qquad\mathrm{(D)}55555\qquad\mathrm{(E)}56432.8$

Solution 1

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 $4! = 24$ ways to arrange the other numbers, so each number appears in each spot $24$ times. Therefore, the sum of all such numbers is $24 \times (1+3+5+7+8) \times (11111) = 24 \times 24 \times 11111 = 6399936.$ Since there are $5! = 120$ such numbers, we divide $6399936 \div 120$ to get $\boxed{\mathrm{(C)}53332.8}$

Solution 2

We can first solve for the mean for the digits 1, 3, 5, 7, and 9 since each is 2 away from each other. The mean of the numbers than can be solved using these digits is $55555$. The total amount of numbers that can be formed using these digits is $4! =120$. The sum of these numbers is $55555(120) = 6666600$. Now we can find out the total value that was gained by replacing the 8 with a 9. We can start how be calculating the gain when the 8 was in the ones digit. Since there are $4! = 24$ numbers with the 8 in the ones digit and 1 was gain from each of them, 24 is the number gained. Then, we repeat this with the tens, hundreds, thousands, and ten thousands place, leading to a total of $24+240+2400+24000+240000=266664$ as the total amount that was gained. Subtract this amount from the sum of the digits using the 9 instead of the 8 to get $6666600-266664=6399936$. Finally, we divide this by 120 to get the average. $6399936/120= \boxed{\mathrm{(C)}53332.8}$

See Also

 2005 AMC 10B (Problems • Answer Key • Resources) Preceded byProblem 19 Followed byProblem 21 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

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