Difference between revisions of "2001 AMC 12 Problems/Problem 12"
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The second case is more or less the same, except we are applying <math>3</math> and <math>4</math> to <math>5</math>. We must find the cases where the first case over counts multiples of five. Utilizing the floor function again on the fractions <math>\lfloor \frac{2001}{3*5} \rfloor</math>, <math>\lfloor \frac{2001}{4*5} \rfloor</math>, and <math>\lfloor \frac{2001}{3*4*5} \rfloor</math> yields the numbers <math>133</math>, <math>100</math>, and <math>33</math>. The first two numbers counted all the numbers that were multiples of either four with five or three with five less than <math>2001</math>. The third counted the overlapping cases, which we must subtract from the sum of the first two. We do this to reach <math>200</math>. Subtracting this number from the original <math>1001</math> numbers procures <math>\boxed{\textbf{(B)}\ 801}</math>. | The second case is more or less the same, except we are applying <math>3</math> and <math>4</math> to <math>5</math>. We must find the cases where the first case over counts multiples of five. Utilizing the floor function again on the fractions <math>\lfloor \frac{2001}{3*5} \rfloor</math>, <math>\lfloor \frac{2001}{4*5} \rfloor</math>, and <math>\lfloor \frac{2001}{3*4*5} \rfloor</math> yields the numbers <math>133</math>, <math>100</math>, and <math>33</math>. The first two numbers counted all the numbers that were multiples of either four with five or three with five less than <math>2001</math>. The third counted the overlapping cases, which we must subtract from the sum of the first two. We do this to reach <math>200</math>. Subtracting this number from the original <math>1001</math> numbers procures <math>\boxed{\textbf{(B)}\ 801}</math>. | ||
− | ==Solution 3== | + | ====Solution 3==== |
First find the number of such numbers between 1 and 2000 (inclusive) and then add one to this result because 2001 is a multiple of 3. | First find the number of such numbers between 1 and 2000 (inclusive) and then add one to this result because 2001 is a multiple of 3. | ||
Revision as of 12:16, 13 June 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 1
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.
Solution 2
We can solve this problem by finding the cases where the number is divisible by or , then subtract from the cases where none of those cases divide . To solve the ways the numbers divide or we find the cases where a number is divisible by and as separate cases. We apply the floor function to every case to get , , and . The first two floor functions were for calculating the number of individual cases for and . The third case was to find any overlapping numbers. The numbers were , , and , respectively. We add the first two terms and subtract the third to get . The first case is finished.
The second case is more or less the same, except we are applying and to . We must find the cases where the first case over counts multiples of five. Utilizing the floor function again on the fractions , , and yields the numbers , , and . The first two numbers counted all the numbers that were multiples of either four with five or three with five less than . The third counted the overlapping cases, which we must subtract from the sum of the first two. We do this to reach . Subtracting this number from the original numbers procures .
Solution 3
First find the number of such numbers between 1 and 2000 (inclusive) and then add one to this result because 2001 is a multiple of 3.
There are numbers that are not multiples of . are not multiples of or , so numbers are.
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 25 |
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.