Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2005 AMC 10B Problems/Problem 13"

(→‎Solution)
(→‎Solution 2)
 
(16 intermediate revisions by 10 users not shown)
Line 2: Line 2:
 
How many numbers between <math>1</math> and <math>2005</math> are integer multiples of <math>3</math> or <math>4</math> but not <math>12</math>?
 
How many numbers between <math>1</math> and <math>2005</math> are integer multiples of <math>3</math> or <math>4</math> but not <math>12</math>?
  
−
<math>\mathrm{(A)} 501 \qquad \mathrm{(B)} 668 \qquad \mathrm{(C)} 835 \qquad \mathrm{(D)} 1002 \qquad \mathrm{(E)} 1169 </math>
+
<math>\textbf{(A) } 501 \qquad \textbf{(B) } 668 \qquad \textbf{(C) } 835 \qquad \textbf{(D) } 1002 \qquad \textbf{(E) } 1169 </math>
−
== Solution ==
+
 
−
We can use the [[Principle of Inclusion-Exclusion]] to solve the problem as follows: We can count the number of multiples of <math>3</math> that are less than <math>2005</math>, add the number of multiples of <math>4</math> that are less than <math>2005</math>, and subtract the number of multiples of <math>12</math> twice that are less than <math>2005</math> (since those are counted twice in each of the <math>3</math> and <math>4</math> cases). Calculating, we get <math>\left\lfloor\dfrac{2005}{3}\right\rfloor+\left\lfloor\dfrac{2005}{4}\right\rfloor-2*\left\lfloor\dfrac{2005}{12}\right\rfloor=668+501-334=\boxed{\mathrm{(c)}\ 835}</math> (where <math>\lfloor x \rfloor</math> denotes the [[floor function]]).
+
== Solution 1 ==
 +
 
 +
To find the multiples of <math>3</math> or <math>4</math> but not <math>12</math>, you need to find the number of multiples of <math>3</math> and <math>4</math>, and then subtract twice the number of multiples of <math>12</math>, because you overcount and do not want to include them. The multiples of <math>3</math> are <math>\frac{2005}{3} = 668\text{ }R1.</math> The multiples of <math>4</math> are <math>\frac{2005}{4} = 501 \text{ }R1</math>. The multiples of <math>12</math> are <math>\frac{2005}{12} = 167\text{ }R1.</math> So, the answer is <math>668+501-167-167 = \boxed{\textbf{(C) } 835}</math>
 +
 
 +
== Solution 2 ==
 +
From <math>1</math>-<math>12</math>, the multiples of <math>3</math> or <math>4</math> but not <math>12</math> are <math>3, 4, 6, 8, </math>and <math>9</math>, a total of five numbers. Since <math>\frac{5}{12}</math> of positive integers are multiples of <math>3</math> or <math>4</math> but not <math>12</math> from <math>1</math>-<math>12</math>, the answer is approximately <math>\frac{5}{12} \cdot 2005</math> = <math>\boxed{\textbf{(C) }835}</math>
  
 
== See Also ==
 
== See Also ==
 
{{AMC10 box|year=2005|ab=B|num-b=12|num-a=14}}
 
{{AMC10 box|year=2005|ab=B|num-b=12|num-a=14}}
 +
{{MAA Notice}}

Latest revision as of 13:44, 15 December 2021

Problem

How many numbers between $1$ and $2005$ are integer multiples of $3$ or $4$ but not $12$?

$\textbf{(A) } 501 \qquad \textbf{(B) } 668 \qquad \textbf{(C) } 835 \qquad \textbf{(D) } 1002 \qquad \textbf{(E) } 1169$

Solution 1

To find the multiples of $3$ or $4$ but not $12$, you need to find the number of multiples of $3$ and $4$, and then subtract twice the number of multiples of $12$, because you overcount and do not want to include them. The multiples of $3$ are $\frac{2005}{3} = 668\text{ }R1.$ The multiples of $4$ are $\frac{2005}{4} = 501 \text{ }R1$. The multiples of $12$ are $\frac{2005}{12} = 167\text{ }R1.$ So, the answer is $668+501-167-167 = \boxed{\textbf{(C) } 835}$

Solution 2

From $1$-$12$, the multiples of $3$ or $4$ but not $12$ are $3, 4, 6, 8,$and $9$, a total of five numbers. Since $\frac{5}{12}$ of positive integers are multiples of $3$ or $4$ but not $12$ from $1$-$12$, the answer is approximately $\frac{5}{12} \cdot 2005$ = $\boxed{\textbf{(C) }835}$

See Also

2005 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_10B_Problems/Problem_13&oldid=167762"