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

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(wikified and linked to "complementary counting")
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== Solution ==
 
== Solution ==
  
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.  
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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.  
  
 
Total # of 4-digit integers: <math>9 * 10 * 10 * 10 = 9000</math>
 
Total # of 4-digit integers: <math>9 * 10 * 10 * 10 = 9000</math>
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== See Also ==
 
== See Also ==
 
*[[2006 AMC 10A Problems]]
 
*[[2006 AMC 10A Problems]]
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* [[Complementary counting]]

Revision as of 11:03, 23 July 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 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.

Total # of 4-digit integers: $9 * 10 * 10 * 10 = 9000$

Total # of 4-digit integers w/o 2 or 3: $7 * 8 * 8 * 8 = 3584$

Therefore, the total number of positive 4-digit integers that have at least one 2 or 3 in it equals: 9000-3584=5416 (E)


See Also

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