Difference between revisions of "2005 AMC 10B Problems/Problem 20"
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<math> \textbf{(A) }48000\qquad\textbf{(B) }49999.5\qquad\textbf{(C) }53332.8\qquad\textbf{(D) }55555\qquad\textbf{(E) }56432.8 </math> | <math> \textbf{(A) }48000\qquad\textbf{(B) }49999.5\qquad\textbf{(C) }53332.8\qquad\textbf{(D) }55555\qquad\textbf{(E) }56432.8 </math> | ||
| − | == Solution 1 == | + | == Solution 1 (Expectation) == |
| − | + | The average of all valid numbers is simply the expected value of a randomly chosen valid number. In other words, it is <math>E[\text{ten thousands digit}]\cdot 10^4+E[\text{thousands digit}]\cdot 10^3+\cdots+E[\text{ones digit}]\cdot 1</math>. However, every digit is equally likely to be chosen for every place, so we can simplify this to <math>\frac{1+3+5+7+8}{5}\cdot(10^4+10^3+10^2+10^1+1)</math> <math>=4.8\cdot11111 = \boxed{(C) 53332.8}</math>. | |
| − | ==Solution 2== | + | == Solution 2 == |
| − | We | + | 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 <math> 4! = 24 </math> ways to arrange the other numbers, so each number appears in each spot <math> 24 </math> times. Therefore, the sum of all such numbers is <math> 24 \times (1+3+5+7+8) \times (11111) = 24 \times 24 \times 11111 = 6399936. </math> Since there are <math> 5! = 120 </math> such numbers, we divide <math> 6399936 \div 120 </math> to get <math> \boxed{\textbf{(C) }53332.8} </math> |
| − | == Solution 3 == | + | ==Solution 3== |
| + | We can first solve for the mean for the digits <math>1, 3, 5, 7, </math>and <math>9</math> since each is <math>2</math> away from each other. The mean of the numbers than can be solved using these digits is <math>55555</math>. The total amount of numbers that can be formed using these digits is <math>5! =120</math>. The sum of these numbers is <math>55555(120) = 6666600</math>. Now we can find out the total value that was gained by replacing the <math>8</math> with a <math>9</math>. We can start how be calculating the gain when the <math>8</math> was in the ones digit. Since there are <math>4! = 24</math> numbers with the <math>8</math> in the ones digit and <math>1</math> was gain from each of them, <math>24</math> is the number gained. Then, we repeat this with the tens, hundreds, thousands, and ten thousands place, leading to a total of <math>24+240+2400+24000+240000=266664</math> as the total amount that was gained. Subtract this amount from the sum of the digits using the <math>9</math> instead of the <math>8</math> to get <math>6666600-266664=6399936</math>. Finally, we divide this by <math>120</math> to get the average. <math>\frac{6399936}{120}= \boxed{\textbf{(C) }53332.8} </math> | ||
| + | |||
| + | == Solution 4 == | ||
The average value of the digits is <math>(1 + 3 + 5 + 7 + 8)/5 = 4.8</math>. | The average value of the digits is <math>(1 + 3 + 5 + 7 + 8)/5 = 4.8</math>. | ||
| − | Values occur in every place so <math>4.8 * 11111 = 53332.8</math>. | + | Values occur in every place so <math>4.8 * 11111 = \boxed{\textbf{(C) }53332.8}</math>. |
| + | |||
| + | == Video Solution == | ||
| + | https://www.youtube.com/watch?v=Ha3xtyz5bME ~David | ||
== See Also == | == See Also == | ||
{{AMC10 box|year=2005|ab=B|num-b=19|num-a=21}} | {{AMC10 box|year=2005|ab=B|num-b=19|num-a=21}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 20:48, 1 September 2024
Contents
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, and 8 exactly once?
Solution 1 (Expectation)
The average of all valid numbers is simply the expected value of a randomly chosen valid number. In other words, it is ![$E[\text{ten thousands digit}]\cdot 10^4+E[\text{thousands digit}]\cdot 10^3+\cdots+E[\text{ones digit}]\cdot 1$](http://latex.artofproblemsolving.com/b/e/6/be61f8ac132713d9739411c3eca02e7d7b338fd0.png)
. However, every digit is equally likely to be chosen for every place, so we can simplify this to 
.
Solution 2
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 
Solution 3
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. 
Solution 4
The average value of the digits is 
.
Values occur in every place so 
.
Video Solution
https://www.youtube.com/watch?v=Ha3xtyz5bME ~David
See Also
| 2005 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 19 |
Followed by Problem 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
