Skip to main content

2005 AMC 10B Problem 20

Problem 20 of 25HarderNumber TheoryCounting & Probability

What is the average (mean) of all 55-digit numbers that can be formed by using each of the digits 1,1, 3,3, 5,5, 7,7, and 88 exactly once?

Answer choices

Show solution

Solution

By symmetry, each of the five digits appears equally often in each place, so the average digit in every place is 1+3+5+7+85=4.8. \dfrac{1 + 3 + 5 + 7 + 8}{5} = 4.8. The average number is therefore 4.8(1+10+100+1000+10000)=4.811111=53332.8. \begin{gathered} \small 4.8(1 + 10 + 100 + 1000 + 10000) \\ = 4.8 \cdot 11111 = 53332.8. \end{gathered} Thus, C is the correct answer.

More practice

Concepts: permutations · place value · symmetry

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.