Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 3)
 
Line 4: Line 4:
 
<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 ==
 
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>
 
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 2==
+
==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>
 
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 3 ==
+
== 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>.

Latest revision as of 20:48, 1 September 2024

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?

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

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 $E[\text{ten thousands digit}]\cdot 10^4+E[\text{thousands digit}]\cdot 10^3+\cdots+E[\text{ones digit}]\cdot 1$. However, every digit is equally likely to be chosen for every place, so we can simplify this to $\frac{1+3+5+7+8}{5}\cdot(10^4+10^3+10^2+10^1+1)$ $=4.8\cdot11111 = \boxed{(C) 53332.8}$.

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 $4! = 24$ 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 $\boxed{\textbf{(C) }53332.8}$

Solution 3

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 $5! =120$. 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. $\frac{6399936}{120}= \boxed{\textbf{(C) }53332.8}$

Solution 4

The average value of the digits is $(1 + 3 + 5 + 7 + 8)/5 = 4.8$. Values occur in every place so $4.8 * 11111 = \boxed{\textbf{(C) }53332.8}$.

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_10B_Problems/Problem_20&oldid=226170"