Difference between revisions of "How many times does the digit 9 appear in the list of all integers from 1 to 500? (The number $ 99 $, for example, is counted twice, because $9$ appears two times in it.)"

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{{delete|Although this is a mathematics problem, the title is unbelievably long. You may want to delete this.}}
 
 
How many times does the digit 9 appear in the list of all integers from 1 to 500? (The number <math> 99 </math>, for example, is counted twice, because <math>9</math> appears two times in it.)
 
The answer is.....
 
 
 
 
The easiest approach is to consider how many times 9 can appear in the units place, how many times in the tens place, and how many times in the hundreds place. If we put a 9 in the units place, there are 10 choices for the tens place and 5 choices for the hundreds digit (including 0), for a total of 50 times. Likewise, if we put a 9 in the tens place, there are 10 choices for the units place and 5 choices for the hundreds digit, for a total of 50 times. Since 9 cannot appear in the hundreds digit, there are <math>50+50=\boxed{100}</math> appearances of the digit 9.
 
The easiest approach is to consider how many times 9 can appear in the units place, how many times in the tens place, and how many times in the hundreds place. If we put a 9 in the units place, there are 10 choices for the tens place and 5 choices for the hundreds digit (including 0), for a total of 50 times. Likewise, if we put a 9 in the tens place, there are 10 choices for the units place and 5 choices for the hundreds digit, for a total of 50 times. Since 9 cannot appear in the hundreds digit, there are <math>50+50=\boxed{100}</math> appearances of the digit 9.
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Latest revision as of 15:13, 30 October 2024

The easiest approach is to consider how many times 9 can appear in the units place, how many times in the tens place, and how many times in the hundreds place. If we put a 9 in the units place, there are 10 choices for the tens place and 5 choices for the hundreds digit (including 0), for a total of 50 times. Likewise, if we put a 9 in the tens place, there are 10 choices for the units place and 5 choices for the hundreds digit, for a total of 50 times. Since 9 cannot appear in the hundreds digit, there are $50+50=\boxed{100}$ appearances of the digit 9.

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