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

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== Problem ==
 
== Problem ==
How many four-digit positive integers have at least one digit that is a 2 or a 3?  
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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 ==
 
== Solution ==
 +
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.
  
Since we are asked for the number of positive 4-[[digit]] [[integer]]s 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.
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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).
 
 
Total # of 4-digit integers: <math>9 * 10 * 10 * 10 = 9000</math>
 
  
Total # of 4-digit integers w/o 2 or 3: <math>7 * 8 * 8 * 8 = 3584</math>
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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 it equals: <math>9000-3584=5416 \Rightarrow E </math>
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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>
  
 
== See Also ==
 
== See Also ==

Revision as of 14:58, 16 October 2006

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 $7 \cdot 8 \cdot 8 \cdot 8 = 3584$.

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

See Also

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