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

Revision as of 21:24, 24 October 2021

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 $5$ ?

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

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 $\boxed{\text{(B)}}$ , 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 $792+9=\boxed{801}$ 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 $\left\lfloor \frac{2001}{3*5} \right\rfloor$ , $\left\lfloor \frac{2001}{4*5} \right\rfloor$ , and $\left\lfloor \frac{2001}{3*4*5} \right\rfloor$ 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 $\frac45*2000=1600$ numbers that are not multiples of $5$ . $\frac23*\frac34*1600=800$ 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 $\frac{10}{25}*2000=800$ 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

Video Solutions

https://youtu.be/EXWK7U8uXyk

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

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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