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 "2006 AMC 10A Problems/Problem 21"

(→‎Solution (Complementary Counting))
(→‎Solution (Casework))
Line 18: Line 18:
 
Therefore, the total number of positive <math>4</math>-digit integers that have at least one <math>2</math> or <math>3</math> is <math>9000-3584=\boxed{\textbf{(E) }5416}.</math>
 
Therefore, the total number of positive <math>4</math>-digit integers that have at least one <math>2</math> or <math>3</math> is <math>9000-3584=\boxed{\textbf{(E) }5416}.</math>
  
−
== Solution (Casework)==
+
== Solution 2 (Casework)==
−
We proceed to the cases.
+
We proceed to every case.
  
−
Case <math>1</math>: There is only one <math>2</math> or <math>3</math>. If the <math>2</math> or <math>3</math> is occupying the first digit, we have <math>512</math> arrangements. If the <math>2</math> or <math>3</math> is not occupying the first digit, there are <math>7 \cdot 8^2</math> = <math>448</math> arrangements. Therefore, we have <math>2(448 \cdot 3 + 512) = 3712</math> arrangements.
+
Case <math>1</math>: There is ONLY one <math>2</math> or <math>3</math>. If the <math>2</math> or <math>3</math> is occupying the first digit, we have <math>512</math> arrangements. If the <math>2</math> or <math>3</math> is not occupying the first digit, there are <math>7 \cdot 8^2</math> = <math>448</math> arrangements. Therefore, we have <math>2(448 \cdot 3 + 512) = 3712</math> arrangements.
  
−
Case <math>2</math> : There are Two <math>2</math>s or two <math>3</math>s but not both. If the <math>2</math> or <math>3</math> is occupying the first digit, we have <math>64</math> arrangements. If the <math>2</math> or <math>3</math> is not occupying the first digit, there are <math>56</math> arrangements. There are <math>3</math> ways for the <math>2</math> or the <math>3</math> to be occupying the first digit and <math>3</math> ways for the first digit to be unoccupied. There are <math>2(3 \cdot (56+64))</math> = <math>720</math> arrangements.
+
Case <math>2</math> : There are two <math>2</math>s OR two <math>3</math>s. If the <math>2</math> or <math>3</math> is occupying the first digit, we have <math>64</math> arrangements. If the <math>2</math> or <math>3</math> is not occupying the first digit, there are <math>56</math> arrangements. There are <math>3</math> ways for the <math>2</math> or the <math>3</math> to be occupying the first digit and <math>3</math> ways for the first digit to be unoccupied. There are <math>2(3 \cdot (56+64))</math> = <math>720</math> arrangements.
  
−
Case <math>3</math> : There is one <math>3</math> and one <math>2</math> but no more. If the <math>2</math> or the <math>3</math> is occupying the first digit, we have <math>6</math> types of arrangements of where the <math>2</math> or <math>3</math> is. We also have <math>64</math> different arrangements for the non-<math>2</math> or <math>3</math> digits. We have <math>6 \cdot 64</math> = <math>384</math> arrangements. If the <math>2</math> or the <math>3</math> isn't occupying the first digit, we have <math>6</math> types of arrangements of where the <math>2</math> or <math>3</math> is. We also have <math>56</math> different arrangements for the non-<math>2</math> or <math>3</math> digits. We have <math>6 \cdot 56</math> = <math>336</math> arrangements for this case. We have <math>336 + 384</math> = <math>720</math> total arrangements for this case.
+
Case <math>3</math> : There is ONLY one <math>2</math> and one <math>3</math>. If the <math>2</math> or the <math>3</math> is occupying the first digit, we have <math>6</math> types of arrangements of where the <math>2</math> or <math>3</math> is. We also have <math>64</math> different arrangements for the non-<math>2</math> or <math>3</math> digits. We have <math>6 \cdot 64</math> = <math>384</math> arrangements. If the <math>2</math> or the <math>3</math> isn't occupying the first digit, we have <math>6</math> types of arrangements of where the <math>2</math> or <math>3</math> is. We also have <math>56</math> different arrangements for the non-<math>2</math> or <math>3</math> digits. We have <math>6 \cdot 56</math> = <math>336</math> arrangements for this case. We have <math>336 + 384</math> = <math>720</math> total arrangements for this case.
  
−
Notice that we already counted <math>3712 + 720 + 720 = 5152</math> cases and we still have a lot of cases left over to count. This is already larger than the second largest answer choice, and therefore, our answer is <math>\boxed{(E)5416}</math>
+
Notice that we already counted <math>3712 + 720 + 720 = 5152</math> cases and we still have a lot of cases left over to count. This is already larger than the second largest answer choice, and therefore, our answer is <math>\boxed{\textbf{(E) }5416}</math>.
  
 
~Arcticturn
 
~Arcticturn

Revision as of 08:40, 19 December 2021

Problem

How many four-digit positive integers have at least one digit that is a $2$ or a $3$?

$\textbf{(A) } 2439\qquad\textbf{(B) } 4096\qquad\textbf{(C) } 4903\qquad\textbf{(D) } 4904\qquad\textbf{(E) } 5416$

Video Solution

https://youtu.be/0W3VmFp55cM?t=3291

~ pi_is_3.14

Solution 1 (Complementary Counting)

Since we are asked for the number of positive $4$-digit integers with at least $2$ or $3$ in it, we can find this by finding the total number of $4$-digit integers and subtracting off those which do not have any $2$s or $3$s as digits.

The total number of $4$-digit integers is $9 \cdot 10 \cdot 10 \cdot 10 = 9000$, since we have $10$ choices for each digit except the first (which can't be $0$).

Similarly, the total number of $4$-digit integers without any $2$ or $3$ is $7 \cdot 8 \cdot 8 \cdot 8 ={3584}$.

Therefore, the total number of positive $4$-digit integers that have at least one $2$ or $3$ is $9000-3584=\boxed{\textbf{(E) }5416}.$

Solution 2 (Casework)

We proceed to every case.

Case $1$: There is ONLY one $2$ or $3$. If the $2$ or $3$ is occupying the first digit, we have $512$ arrangements. If the $2$ or $3$ is not occupying the first digit, there are $7 \cdot 8^2$ = $448$ arrangements. Therefore, we have $2(448 \cdot 3 + 512) = 3712$ arrangements.

Case $2$ : There are two $2$s OR two $3$s. If the $2$ or $3$ is occupying the first digit, we have $64$ arrangements. If the $2$ or $3$ is not occupying the first digit, there are $56$ arrangements. There are $3$ ways for the $2$ or the $3$ to be occupying the first digit and $3$ ways for the first digit to be unoccupied. There are $2(3 \cdot (56+64))$ = $720$ arrangements.

Case $3$ : There is ONLY one $2$ and one $3$. If the $2$ or the $3$ is occupying the first digit, we have $6$ types of arrangements of where the $2$ or $3$ is. We also have $64$ different arrangements for the non-$2$ or $3$ digits. We have $6 \cdot 64$ = $384$ arrangements. If the $2$ or the $3$ isn't occupying the first digit, we have $6$ types of arrangements of where the $2$ or $3$ is. We also have $56$ different arrangements for the non-$2$ or $3$ digits. We have $6 \cdot 56$ = $336$ arrangements for this case. We have $336 + 384$ = $720$ total arrangements for this case.

Notice that we already counted $3712 + 720 + 720 = 5152$ cases and we still have a lot of cases left over to count. This is already larger than the second largest answer choice, and therefore, our answer is $\boxed{\textbf{(E) }5416}$.

~Arcticturn

See also

2006 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 20 		Followed by
Problem 22
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=2006_AMC_10A_Problems/Problem_21&oldid=168009"