Difference between revisions of "2001 AMC 12 Problems/Problem 12"
(New page: == Problem == How many positive integers not exceeding <math>2001</math> are multiple of <math>3</math> or <math>4</math> but not <math>5</math>? <math> \text{(A) }768 \qquad \text{(B) }...) |
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Revision as of 22:52, 16 March 2011
Problem
How many positive integers not exceeding are multiple 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.
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 |