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

(Solution 3)
(Solution 2)
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This is correct.
 
This is correct.
 
===Solution 2===
 
===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.
+
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>\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)}\ 801}</math>.
+
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*5} \right\rfloor</math>, <math>\left\lfloor \frac{2001}{4*5} \right\rfloor</math>, and <math>\left\lfloor \frac{2001}{3*4*5} \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===
 
===Solution 3===

Revision as of 13:18, 13 June 2017

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 $5$?

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

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 numbers 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. $800+1=\boxed{\textbf{(B)}\ 801}$

See Also

2001 AMC 12 (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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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