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

Latest revision as of 13:34, 17 August 2023

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

Video Solution

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

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.

2003 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 7: / Line 7: @@
 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>(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>.
-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>
+==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 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 Problems]]
+{{AMC10 box|year=2003|ab=A|num-b=14|num-a=16}}
-*[[2003 AMC 10A Problems/Problem 14|Previous Problem]]
-*[[2003 AMC 10A Problems/Problem 16|Next Problem]]
 [[Category:Introductory Number Theory Problems]]
+{{MAA Notice}}

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