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

(Solution)
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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).
 
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>.
+
Similarly, the total number of 4-digit integers without any 2 or 3 is <math>7 \cdot 8 \cdot 8 \cdot 8 =[boxed] 3584[\boxed]</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=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=5416 \Longrightarrow \mathrm{(E)} </math>

Revision as of 10:55, 23 February 2017

Problem

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

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

Solution

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 $7 \cdot 8 \cdot 8 \cdot 8 =[boxed] 3584[\boxed]$.

Therefore, the total number of positive 4-digit integers that have at least one 2 or 3 in their decimal representation is $9000-3584=5416 \Longrightarrow \mathrm{(E)}$

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

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