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

(Solution (Complementary Counting))
(Solution (Casework))
Line 17: Line 17:
  
 
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>
 
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>
 
==Solution (Casework)==
 
  
 
== See also ==
 
== See also ==

Revision as of 20:00, 25 October 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 $9000-3584=\boxed{5416} \Longrightarrow \mathrm{(E)}$

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