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

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== Problem ==
 
== Problem ==
How many four-[[digit]] [[positive integer]]s have at least one digit that is a <math>2</math> or a <math>3</math>?
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How many four-digit positive integers have at least one digit that is a <math>2</math> or a <math>3</math>?
  
 
<math>\textbf{(A) } 2439\qquad\textbf{(B) } 4096\qquad\textbf{(C) } 4903\qquad\textbf{(D) } 4904\qquad\textbf{(E) } 5416</math>
 
<math>\textbf{(A) } 2439\qquad\textbf{(B) } 4096\qquad\textbf{(C) } 4903\qquad\textbf{(D) } 4904\qquad\textbf{(E) } 5416</math>

Revision as of 09:14, 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 (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 2s or 3s 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 in their decimal representation is $9000-3584=\boxed{5416} \Longrightarrow \mathrm{(E)}$

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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