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Difference between revisions of "2001 AMC 12 Problems/Problem 12"

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Similar to some of the above solutions.  
 
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 60 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>.
 
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 60 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==
 
==Video Solutions==

Revision as of 13:12, 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.

Problem

How many positive integers not exceeding $2001$ are multiples of $3$ or $4$ but not $5$?

$\textbf{(A) }768 \qquad \textbf{(B) }801 \qquad \textbf{(C) }934 \qquad \textbf{(D) }1067 \qquad \textbf{(E) }1167$

Solutions

Solution 1

Out of the numbers $1$ to $12$ four are divisible by $3$ and three by $4$, counting $12$ twice. Hence $6$ out of these $12$ numbers are multiples of $3$ or $4$.

The same is obviously true for the numbers $12k+1$ to $12k+12$ for any positive integer $k$.

Hence out of the numbers $1$ to $60=5\cdot 12$ there are $5\cdot 6=30$ numbers that are divisible by $3$ or $4$. Out of these $30$, the numbers $15$, $20$, $30$, $40$, $45$ and $60$ are divisible by $5$. Therefore in the set $\{1,\dots,60\}$ there are precisely $30-6=24$ numbers that satisfy all criteria from the problem statement.

Again, the same is obviously true for the set $\{60k+1,\dots,60k+60\}$ for any positive integer $k$.

We have $1980/60 = 33$, hence there are $24\cdot 33 = 792$ good numbers among the numbers $1$ to $1980$. At this point we already know that the only answer that is still possible is $\boxed{\textbf{(B)}}$, as we only have $20$ numbers left.

By examining the remaining $20$ by hand we can easily find out that exactly $9$ of them match all the criteria, giving us $792+9=\boxed{\textbf{(B) }801}$ good numbers. This is correct.

Solution 2

We can solve this problem by finding the cases where the number is divisible by $3$ or $4$, then subtract from the cases where none of those cases divide $5$. To solve the ways the numbers divide $3$ or $4$ we find the cases where a number is divisible by $3$ and $4$ as separate cases. We apply the floor function to every case to get $\left\lfloor \frac{2001}{3} \right\rfloor$, $\left\lfloor \frac{2001}{4} \right\rfloor$, and $\left\lfloor \frac{2001}{12} \right\rfloor$. The first two floor functions were for calculating the number of individual cases for $3$ and $4$. The third case was to find any overlapping numbers. The numbers were $667$, $500$, and $166$, respectively. We add the first two terms and subtract the third to get $1001$. The first case is finished.

The second case is more or less the same, except we are applying $3$ and $4$ to $5$. We must find the cases where the first case over counts multiples of five. Utilizing the floor function again on the fractions $\left\lfloor \frac{2001}{3\cdot5} \right\rfloor$, $\left\lfloor \frac{2001}{4\cdot5} \right\rfloor$, and $\left\lfloor \frac{2001}{3\cdot4\cdot5} \right\rfloor$ yields the numbers $133$, $100$, and $33$. The first two numbers counted all the numbers that were multiples of either four with five or three with five less than $2001$. The third counted the overlapping cases, which we must subtract from the sum of the first two. We do this to reach $200$. Subtracting this number from the original $1001$ numbers procures $\boxed{\textbf{(B)}\ 801}$.

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 $3$.

There are $\frac45\cdot2000=1600$ numbers that are not multiples of $5$. $\frac23\cdot\frac34\cdot1600=800$ are not multiples of $3$ or $4$, so $800$ numbers are. $800+1=\boxed{\textbf{(B)}\ 801}$

Solution 4

Take a good-sized sample of consecutive integers; for example, the first $25$ positive integers. Determine that the numbers $3, 4, 6, 8, 9, 12, 16, 18, 21,$ and $24$ exhibit the properties given in the question. $25$ is a divisor of $2000$, so there are $\frac{10}{25}\cdot2000=800$ numbers satisfying the given conditions between $1$ and $2000$. Since $2001$ is a multiple of $3$, add $1$ to $800$ to get $800+1=\boxed{\textbf{(B)}\ 801}$.

~ mathmagical

Solution 5

By PIE, there are $1001$ numbers that are multiples of $3$ or $4$ and less than or equal to $2001$. $80\%$ of them will not be divisible by $5$, and by far the closest number to $80\%$ of $1001$ is $\boxed{\textbf{(B)}\ 801}$.

~ Fasolinka

Solution 5

Similar to some of the above solutions. We can divide $2001$ by $3$ and $4$ to find the number of integers divisible by $3$ and $4$. Hence, we find that there are $667$ numbers less than $2001$ that are divisible by $3$, and $500$ numbers that are divisible by $4$. However, we will need to subtract the number of multiples of $15$ from 667 and that of $20$ from $500$, since they're also divisible by 5 which we don't want. There are $133$ + $100$ = $233$ such numbers. Note that during this process, we've subtracted the multiples of $60$ twice because they're divisible by both $15$ and $20$, so we have to add $33$ back to the tally (there are $33$ multiples of 60 that does not exceed $2001$). Lastly, we have to subtract multiples of both $3$ AND $4$ since we only want multiples of either $3$ or $4$. This is tantamount to subtracting the number of multiples of $12$. And there are $166$ such numbers. Let's now collect our numbers and compute the total: $667$ + $500$ - $133$ - $100$ + $33$ - $166$ = $\boxed{\textbf{(B)}\ 801}$.

~ PlainOldNumberTheory

Video Solutions

https://youtu.be/EXWK7U8uXyk

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

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