Difference between revisions of "2001 AMC 12 Problems/Problem 12"
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<math> | <math> | ||
− | \ | + | \textbf{(A) }768 |
\qquad | \qquad | ||
− | \ | + | \textbf{(B) }801 |
\qquad | \qquad | ||
− | \ | + | \textbf{(C) }934 |
\qquad | \qquad | ||
− | \ | + | \textbf{(D) }1067 |
\qquad | \qquad | ||
− | \ | + | \textbf{(E) }1167 |
</math> | </math> | ||
− | == Solution == | + | ==Solutions== |
+ | === 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. | ||
Line 30: | Line 31: | ||
Again, the same is obviously true for the set <math>\{60k+1,\dots,60k+60\}</math> for any positive integer <math>k</math>. | Again, the same is obviously true for the set <math>\{60k+1,\dots,60k+60\}</math> for any positive integer <math>k</math>. | ||
− | We have <math>1980/60 = 33</math>, hence there are <math>24\cdot 33 = 792</math> good numbers among the numbers <math>1</math> to <math>1980</math>. At this point we already know that the only answer that is still possible is <math>\boxed{\ | + | We have <math>1980/60 = 33</math>, hence there are <math>24\cdot 33 = 792</math> good numbers among the numbers <math>1</math> to <math>1980</math>. At this point we already know that the only answer that is still possible is <math>\boxed{\textbf{(B)}}</math>, as we only have <math>20</math> numbers left. |
− | 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{\textbf{(B) }801}</math> good numbers. |
+ | 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>\left\lfloor \frac{2001}{3} \right\rfloor</math>, <math>\left\lfloor \frac{2001}{4} \right\rfloor</math>, and <math>\left\lfloor \frac{2001}{12} \right\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. | ||
+ | |||
+ | 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>\left\lfloor \frac{2001}{3\cdot5} \right\rfloor</math>, <math>\left\lfloor \frac{2001}{4\cdot5} \right\rfloor</math>, and <math>\left\lfloor \frac{2001}{3\cdot4\cdot5} \right\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=== | ||
+ | First find the number of such integers between 1 and 2000 (inclusive) and then add one to this result because 2001 is a multiple of <math>3</math>. | ||
+ | |||
+ | There are <math>\frac45\cdot2000=1600</math> numbers that are not multiples of <math>5</math>. <math>\frac23\cdot\frac34\cdot1600=800</math> are not multiples of <math>3</math> or <math>4</math>, so <math>800</math> numbers are. <math>800+1=\boxed{\textbf{(B)}\ 801}</math> | ||
+ | |||
+ | ===Solution 4=== | ||
+ | Take a good-sized sample of consecutive integers; for example, the first <math>25</math> positive integers. Determine that the numbers <math>3, 4, 6, 8, 9, 12, 16, 18, 21,</math> and <math>24</math> exhibit the properties given in the question. <math>25</math> is a divisor of <math>2000</math>, so there are <math>\frac{10}{25}\cdot2000=800</math> numbers satisfying the given conditions between <math>1</math> and <math>2000</math>. Since <math>2001</math> is a multiple of <math>3</math>, add <math>1</math> to <math>800</math> to get <math>800+1=\boxed{\textbf{(B)}\ 801}</math>. | ||
+ | |||
+ | ~ mathmagical | ||
+ | |||
+ | ===Solution 5=== | ||
+ | By PIE, there are <math>1001</math> numbers that are multiples of <math>3</math> or <math>4</math> and less than or equal to <math>2001</math>. <math>80\%</math> of them will not be divisible by <math>5</math>, and by far the closest number to <math>80\%</math> of <math>1001</math> is <math>\boxed{\textbf{(B)}\ 801}</math>. | ||
+ | |||
+ | ~ Fasolinka | ||
+ | |||
+ | === Solution 5=== | ||
+ | Similar to some of the above solutions. | ||
+ | We can divide <math>2001</math> by <math>3</math> and <math>4</math> to find the number of integers divisible by <math>3</math> and <math>4</math>. Hence, we find that there are <math>667</math> numbers less than <math>2001</math> that are divisible by <math>3</math>, and <math>500</math> numbers that are divisible by <math>4</math>. However, we will need to subtract the number of multiples of <math>15</math> from 667 and that of <math>20</math> from <math>500</math>, since they're also divisible by 5 which we don't want. There are <math>133</math> + <math>100</math> = <math>233</math> such numbers. Note that during this process, we've subtracted the multiples of <math>60</math> twice because they're divisible by both <math>15</math> and <math>20</math>, so we have to add <math>33</math> back to the tally (there are <math>33</math> multiples of <math>60</math> that does not exceed <math>2001</math>). Lastly, we have to subtract multiples of both <math>3</math> AND <math>4</math> since we only want multiples of either <math>3</math> or <math>4</math>. This is tantamount to subtracting the number of multiples of <math>12</math>. And there are <math>166</math> such numbers. Let's now collect our numbers and compute the total: <math>667</math> + <math>500</math> - <math>133</math> - <math>100</math> + <math>33</math> - <math>166</math> = <math>\boxed{\textbf{(B)}\ 801}</math>. | ||
+ | |||
+ | ~ PlainOldNumberTheory | ||
+ | |||
+ | ==Video Solutions== | ||
+ | https://youtu.be/EXWK7U8uXyk | ||
+ | |||
+ | https://www.youtube.com/watch?v=XHmKu-ZoRxI&feature=youtu.be | ||
== See Also == | == See Also == | ||
{{AMC12 box|year=2001|num-b=11|num-a=13}} | {{AMC12 box|year=2001|num-b=11|num-a=13}} | ||
− | {{AMC10 box|year=2001|num-b= | + | {{AMC10 box|year=2001|num-b=25|after=Last Question}} |
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 14:14, 1 May 2022
- 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 ?
Solutions
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 integers between 1 and 2000 (inclusive) and then add one to this result because 2001 is a multiple of .
There are numbers that are not multiples of . are not multiples of or , so numbers are.
Solution 4
Take a good-sized sample of consecutive integers; for example, the first positive integers. Determine that the numbers and exhibit the properties given in the question. is a divisor of , so there are numbers satisfying the given conditions between and . Since is a multiple of , add to to get .
~ mathmagical
Solution 5
By PIE, there are numbers that are multiples of or and less than or equal to . of them will not be divisible by , and by far the closest number to of is .
~ Fasolinka
Solution 5
Similar to some of the above solutions. We can divide by and to find the number of integers divisible by and . Hence, we find that there are numbers less than that are divisible by , and numbers that are divisible by . However, we will need to subtract the number of multiples of from 667 and that of from , since they're also divisible by 5 which we don't want. There are + = such numbers. Note that during this process, we've subtracted the multiples of twice because they're divisible by both and , so we have to add back to the tally (there are multiples of that does not exceed ). Lastly, we have to subtract multiples of both AND since we only want multiples of either or . This is tantamount to subtracting the number of multiples of . And there are such numbers. Let's now collect our numbers and compute the total: + - - + - = .
~ PlainOldNumberTheory
Video Solutions
https://www.youtube.com/watch?v=XHmKu-ZoRxI&feature=youtu.be
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.