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

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There are <math>100</math> integers in the set.  
 
There are <math>100</math> integers in the set.  
  
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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.  
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Since every <math>2^{\text{nd}}</math> integer is divisible by <math>2</math>, there are <math>\left\lfloor\frac{100}{2}\right\rfloor = 50</math> integers divisible by <math>2</math> in the set.  
  
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To be divisible by both <math>2</math> and <math>3</math>, a number must be divisible by <math>(2,3)=6</math>.
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To be divisible by both <math>2</math> and <math>3</math>, a number must be divisible by <math>\operatorname{lcm}(2,3)=6</math>.
  
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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.  
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Since every <math>6^{\text{th}}</math> integer is divisible by <math>6</math>, there are <math>\left\lfloor\frac{100}{6}\right\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>.  
  
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Therefore, the desired probability is <math>\frac{34}{100}=\frac{17}{50}\Rightarrow\boxed{\mathrm{(C)}\ \frac{17}{50}}</math>
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Therefore, the desired probability is <math>\frac{34}{100}=\frac{17}{50}\Rightarrow\boxed{\mathrm{(C)}\ \frac{17}{50}}</math>.
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==Video Solution by WhyMath==
 +
https://youtu.be/UfzS5griBic
 +
 
 +
~savannahsolver
 +
 
 +
==Video Solution==
 +
https://www.youtube.com/watch?v=4IlfkRW660E  ~David
  
 
==Controversy==
 
==Controversy==
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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.
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Due to the wording of the problem statement, it might be taken as "Find the probability that an integer '''''exists''''' in said set that is divisible by <math>2</math> and not <math>3</math>". One example would be <math>2</math>, which is not a multiple of <math>3</math>, and thus the probability would appear to be <math>1</math>. But because <math>1</math> is not an answer choice, we can assume that that was not the intended meaning.
  
 
== See Also ==
 
== See Also ==

Latest revision as of 14:29, 20 June 2025

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 $\left\lfloor\frac{100}{2}\right\rfloor = 50$ integers divisible by $2$ in the set.

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

Since every $6^{\text{th}}$ integer is divisible by $6$, there are $\left\lfloor\frac{100}{6}\right\rfloor = 16$ 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

Video Solution

https://www.youtube.com/watch?v=4IlfkRW660E ~David

Controversy

Due to the wording of the problem statement, it might be taken as "Find the probability that an integer exists in said set that is divisible by $2$ and not $3$". One example would be $2$, which is not a multiple of $3$, and thus the probability would appear to be $1$. But because $1$ is not an answer choice, we can assume that that was not the intended meaning.

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

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