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

Latest revision as of 14:20, 30 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

Solution 6

Similar to @above: Let the function $M_{2001}(n)$ return how many multiples of $n$ are there not exceeding $2001$ . Then we have that the desired number is: $M_{2001}(3)+M_{2001}(4)-M_{2001}(3\cdot 4)-M_{2001}(3 \cdot 5) - M_{2001}(4 \cdot 5)+M_{2001}(3 \cdot 4 \cdot 5)$

Evaluating each of these we get: $667+500-166-133-100+33 = 1100-299 = 801.$

Thus, the answer is $\boxed{\textbf{(B)}\ 801}.$

-ConfidentKoala4

Video Solutions

https://youtu.be/EXWK7U8uXyk

https://www.youtube.com/watch?v=XHmKu-ZoRxI&feature=youtu.be

@@ Line 58: / Line 58: @@
 === 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 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 <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
+=== Solution 6===
+Similar to @above:
+Let the function <math>M_{2001}(n)</math> return how many multiples of <math>n</math> are there not exceeding <math>2001</math>. Then we have that the desired number is:
+<cmath>M_{2001}(3)+M_{2001}(4)-M_{2001}(3\cdot 4)-M_{2001}(3 \cdot 5) - M_{2001}(4 \cdot 5)+M_{2001}(3 \cdot 4 \cdot 5)</cmath>
+Evaluating each of these we get:
+<cmath>667+500-166-133-100+33 = 1100-299 = 801.</cmath>
+Thus, the answer is <math>\boxed{\textbf{(B)}\ 801}.</math>
+-ConfidentKoala4
 ==Video Solutions==

2001 AMC 12 (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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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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Difference between revisions of "2001 AMC 12 Problems/Problem 12"

Latest revision as of 14:20, 30 May 2022

Contents

Problem

Solutions

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 5

Solution 6

Video Solutions

See Also