Difference between revisions of "2001 AMC 12 Problems/Problem 12"
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| − | == Solution == | + | == Solution 1== |
Out of the numbers <math>1</math> to <math>12</math> four are divisible by <math>3</math> and three by <math>4</math>, counting <math>12</math> twice. | Out of the numbers <math>1</math> to <math>12</math> four are divisible by <math>3</math> and three by <math>4</math>, counting <math>12</math> twice. | ||
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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. | This is correct. | ||
| + | ===Solution 2=== | ||
| + | We can solve this problem by finding the cases where the number is divisible by <math>3</math> or <math>4</math>, then subtract from the cases where none of those cases divide <math>5</math>. To solve the ways the numbers divide <math>3</math> or <math>4</math> we find the cases where a number is divisible by <math>3</math> and <math>4</math> as separate cases. We apply the floor function to every case to get <math>\lfloor \frac{2001}{3} \rfloor</math>, <math>\lfloor \frac{2001}{4} \rfloor</math>, and <math>\lfloor \frac{2001}{12} \rfloor</math>. The first two floor functions were for calculating the number of individual cases for <math>3</math> and <math>4</math>. The third case was to find any overlapping numbers. The numbers were <math>667</math>, <math>500</math>, and <math>166</math>, respectively. We add the first two terms and subtract the third to get <math>1001</math>. The first case is finished. | ||
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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)}\ 810}</math>. | ||
== See Also == | == See Also == | ||
Revision as of 17:43, 22 May 2017
- The following problem is from both the 2001 AMC 12 #12 and 2001 AMC 10 #25, so both problems redirect to this page.
Contents
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 
.
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. 
