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 "2003 AMC 10A Problems/Problem 15"
m (→‎Solution)
Line 7: Line 7:
 
There are <math>100</math> integers in the set.  
 
There are <math>100</math> integers in the set.  
  
−
Since every 2nd integer is divisible by <math>2</math>, there are <math>\lfloor\frac{100}{2}\rfloor=50</math> integers divisible by <math>2</math> in the set.  
+
Since every <math>2^{\text{nd}}</math> integer is divisible by <math>2</math>, there are <math>\lfloor\frac{100}{2}\rfloor=50</math> integers divisible by <math>2</math> in the set.  
  
−
To be divisible by both <math>2</math> and <math>3</math>, a number must be divisible by <math>lcm(2,3)=6</math>.
+
To be divisible by both <math>2</math> and <math>3</math>, a number must be divisible by <math>\text{lcm}(2,3)=6</math>.
  
−
Since every 6th integer is divisible by <math>6</math>, there are <math>\lfloor\frac{100}{6}\rfloor=16</math> integers divisible by both <math>2</math> and <math>3</math> in the set.  
+
Since every <math>6^{\text{th}}</math> integer is divisible by <math>6</math>, there are <math>\lfloor\frac{100}{6}\rfloor=16</math> integers divisible by both <math>2</math> and <math>3</math> in the set.  
  
 
So there are <math>50-16=34</math> integers in this set that are divisible by <math>2</math> and not divisible by <math>3</math>.  
 
So there are <math>50-16=34</math> integers in this set that are divisible by <math>2</math> and not divisible by <math>3</math>.  
  
−
Therefore, the desired probability is <math>\frac{34}{100}=\frac{17}{50} \Rightarrow C</math>
+
Therefore, the desired probability is <math>\frac{34}{100}=\frac{17}{50}\Rightarrow\boxed{\mathrm{(C)}\ \frac{17}{50}}</math>
  
 
==Controversy==
 
==Controversy==

Revision as of 22:40, 16 July 2014

Problem

What is the probability that an integer in the set $\{1,2,3,...,100\}$ is divisible by $2$ and not divisible by $3$?

$\mathrm{(A) \ } \frac{1}{6}\qquad \mathrm{(B) \ } \frac{33}{100}\qquad \mathrm{(C) \ } \frac{17}{50}\qquad \mathrm{(D) \ } \frac{1}{2}\qquad \mathrm{(E) \ } \frac{18}{25}$

Solution

There are $100$ integers in the set.

Since every $2^{\text{nd}}$ integer is divisible by $2$, there are $\lfloor\frac{100}{2}\rfloor=50$ integers divisible by $2$ in the set.

To be divisible by both $2$ and $3$, a number must be divisible by $\text{lcm}(2,3)=6$.

Since every $6^{\text{th}}$ integer is divisible by $6$, there are $\lfloor\frac{100}{6}\rfloor=16$ integers divisible by both $2$ and $3$ in the set.

So there are $50-16=34$ integers in this set that are divisible by $2$ and not divisible by $3$.

Therefore, the desired probability is $\frac{34}{100}=\frac{17}{50}\Rightarrow\boxed{\mathrm{(C)}\ \frac{17}{50}}$

Controversy

Due to the wording of the question, it may be taken as "Find the probability that an integer in said set is divisible by 2 and not 3 EXISTS". One example would be 2, which is not a multiple of 3, thus the probability is 1.

See Also

2003 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
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=2003_AMC_10A_Problems/Problem_15&oldid=62525"