2006 AMC 10A Problems/Problem 21

Revision as of 20:16, 25 October 2021 by Arcticturn (talk | contribs) (Solution (Casework))

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$ (Error compiling LaTeX. Unknown error_msg)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$ (Error compiling LaTeX. Unknown error_msg)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$ (Error compiling LaTeX. Unknown error_msg)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) }

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png