Difference between revisions of "2006 AMC 10A Problems/Problem 21"

(Problem)
(Solution (Complementary Counting))
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~ pi_is_3.14
 
~ pi_is_3.14
  
== Solution (Complementary Counting) ==
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== Solution 1 (Complementary Counting) ==
Since we are asked for the number of positive 4-digit [[integer]]s 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 2s or 3s as digits.  
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Since we are asked for the number of positive <math>4</math>-digit integers with at least <math>2</math> or <math>3</math> in it, we can find this by finding the total number of <math>4</math>-digit integers and subtracting off those which do not have any <math>2</math>s or <math>3</math>s as digits.  
  
The total number of 4-digit integers is <math>9 \cdot 10 \cdot 10 \cdot 10 = 9000</math>, since we have 10 choices for each digit except the first (which can't be 0).
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The total number of <math>4</math>-digit integers is <math>9 \cdot 10 \cdot 10 \cdot 10 = 9000</math>, since we have <math>10</math> choices for each digit except the first (which can't be <math>0</math>).
  
Similarly, the total number of 4-digit integers without any 2 or 3 is <math>7 \cdot 8 \cdot 8 \cdot 8 ={3584}</math>.
+
Similarly, the total number of <math>4</math>-digit integers without any <math>2</math> or <math>3</math> is <math>7 \cdot 8 \cdot 8 \cdot 8 ={3584}</math>.
  
Therefore, the total number of positive 4-digit integers that have at least one 2 or 3 in their decimal representation is <math>9000-3584=\boxed{5416} \Longrightarrow \mathrm{(E)} </math>
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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 (Casework)==

Revision as of 09:16, 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 (Casework)

We proceed to the cases.

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 but not both. 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 one $3$ and one $2$ but no more. 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 $\boxed{(E)5416}$

~Arcticturn

See also

2006 AMC 10A (ProblemsAnswer KeyResources)
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

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