Difference between revisions of "2003 AMC 10A Problems/Problem 15"

(Controversy)
Line 16: Line 16:
  
 
Therefore, the desired probability is <math>\frac{34}{100}=\frac{17}{50}\Rightarrow\boxed{\mathrm{(C)}\ \frac{17}{50}}</math>
 
Therefore, the desired probability is <math>\frac{34}{100}=\frac{17}{50}\Rightarrow\boxed{\mathrm{(C)}\ \frac{17}{50}}</math>
 +
 +
==Video Solution by WhyMath==
 +
https://youtu.be/UfzS5griBic
 +
 +
~savannahsolver
  
 
==Controversy==
 
==Controversy==

Revision as of 13:19, 14 June 2021

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 $(2,3)=6$.

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 $\frac{34}{100}=\frac{17}{50}\Rightarrow\boxed{\mathrm{(C)}\ \frac{17}{50}}$

Video Solution by WhyMath

https://youtu.be/UfzS5griBic

~savannahsolver

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. But because 1 is not an option, we can assume that it was not meant like that.

See Also

2003 AMC 10A (ProblemsAnswer KeyResources)
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