Difference between revisions of "2001 AMC 12 Problems/Problem 12"
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By examining the remaining <math>20</math> by hand we can easily find out that exactly <math>9</math> of them match all the criteria, giving us <math>792+9=\boxed{801}</math> good numbers. | By examining the remaining <math>20</math> by hand we can easily find out that exactly <math>9</math> of them match all the criteria, giving us <math>792+9=\boxed{801}</math> good numbers. | ||
+ | This is correct. | ||
== See Also == | == See Also == |
Revision as of 18:08, 27 April 2017
- The following problem is from both the 2001 AMC 12 #12 and 2001 AMC 10 #25, so both problems redirect to this page.
Problem
How many positive integers not exceeding are multiples of or but not ?
Solution
Out of the numbers to four are divisible by and three by , counting twice. Hence out of these numbers are multiples of or .
The same is obviously true for the numbers to for any positive integer .
Hence out of the numbers to there are numbers that are divisible by or . Out of these , the numbers , , , , and are divisible by . Therefore in the set there are precisely numbers that satisfy all criteria from the problem statement.
Again, the same is obviously true for the set for any positive integer .
We have , hence there are good numbers among the numbers to . At this point we already know that the only answer that is still possible is , as we only have numbers left.
By examining the remaining by hand we can easily find out that exactly of them match all the criteria, giving us good numbers. This is correct.
See Also
2001 AMC 12 (Problems • Answer Key • Resources) | |
Preceded by Problem 11 |
Followed by Problem 13 |
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 12 Problems and Solutions |
2001 AMC 10 (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last Question | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.