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

(See Also)
(Solution (Complementary Counting))
 
(14 intermediate revisions by 10 users not shown)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
How many four-digit positive integers have at least one digit that is a 2 or a 3?  
+
How many four-[[digit]] [[positive integer]]s have at least one digit that is a 2 or a 3?  
  
 
<math>\mathrm{(A) \ } 2439\qquad\mathrm{(B) \ } 4096\qquad\mathrm{(C) \ } 4903\qquad\mathrm{(D) \ } 4904\qquad\mathrm{(E) \ } 5416\qquad</math>
 
<math>\mathrm{(A) \ } 2439\qquad\mathrm{(B) \ } 4096\qquad\mathrm{(C) \ } 4903\qquad\mathrm{(D) \ } 4904\qquad\mathrm{(E) \ } 5416\qquad</math>
== Solution ==
 
== See Also ==
 
*[[2006 AMC 10A Problems]]
 
  
Since we are asked for the number of positive 4-digit integers with AT LEAST ONE 2 or 3 in it, we can find this by finding the number of 4-digit + integers that DO NOT contain any 2 or 3.  
+
== Video Solution ==
 +
https://youtu.be/0W3VmFp55cM?t=3291
  
Total # of 4-digit integers: <math>9 * 10 * 10 * 10 = 9000</math>
+
~ pi_is_3.14
  
Total # of 4-digit integers w/o 2 or 3: <math>7 * 8 * 8 * 8 = 3584</math>
+
== Solution (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.
  
Therefore, the total number of positive 4-digit integers that have at least one 2 or 3 in it equals: 9000-3584=5416 (E)
+
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).
 +
 
 +
Similarly, the total number of 4-digit integers without any 2 or 3 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>
 +
 
 +
== See also ==
 +
{{AMC10 box|year=2006|ab=A|num-b=20|num-a=22}}
 +
 
 +
[[Category:Introductory Combinatorics Problems]]
 +
{{MAA Notice}}

Latest revision as of 22:26, 16 January 2021

Problem

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

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

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

Invalid username
Login to AoPS