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

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== See Also == | == See Also == | ||

*[[2006 AMC 10A Problems]] | *[[2006 AMC 10A Problems]] | ||

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+ | 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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+ | Total # of 4-digit integers: <math>9 * 10 * 10 * 10 = 9000</math> | ||

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+ | Total # of 4-digit integers w/o 2 or 3: <math>7 * 8 * 8 * 8 = 3584</math> | ||

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

## Revision as of 02:57, 23 July 2006

## Problem

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

## Solution

## See Also

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:

Total # of 4-digit integers w/o 2 or 3:

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