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2019 Mock AMC 10B Problems/Problem 3: Difference between revisions

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==Solution==
==Solution==
After trying each option we have  
After trying each option we have  
<cmath></cmath>A) <math>3^\frac{2018}{3}</math> which is irrational as 2018 is not divisible by 3
 
<cmath></cmath>B) <math>3^\frac{2019}{2}</math> which is irrational as 2019 isn't divisible by 2
A) <math>3^\frac{2018}{3}</math> which is irrational as 2018 is not divisible by 3
<cmath></cmath>C) <math>3^2+\sqrt{2}^2+6\sqrt{2}</math> which equals <math>11+6\sqrt{2}</math> which is irrational
 
<cmath></cmath>D) <math>(2pi)^2</math> equals <math>4pi^2</math>, which is irrational
B) <math>3^\frac{2019}{2}</math> which is irrational as 2019 isn't divisible by 2
<cmath></cmath>E) <math>(3-\sqrt{2})(3+\sqrt{2})=9-2=7</math> which is rational
 
We have <math>\boxed{\bold{E}}</math> <math>(3-\sqrt{2})(3+\sqrt{2})</math>
C) <math>3^2+\sqrt{2}^2+6\sqrt{2}</math> which equals <math>11+6\sqrt{2}</math> which is irrational
 
D) <math>(2\pi)^2</math> equals <math>4\pi^2</math>, which is irrational
 
E) <math>(3-\sqrt{2})(3+\sqrt{2})=9-2=7</math> which is rational
 
Our answer is <math>\boxed{\textbf{(E) }(3-\sqrt{2})(3+\sqrt{2})}</math>.

Latest revision as of 17:04, 27 October 2025

Problem 3

Which of these numbers is a rational number?


$\textbf{(A) }(\sqrt[3]{3})^{2018} \qquad \textbf{(B) }(\sqrt{3})^{2019} \qquad \textbf{(C) }(3+\sqrt{2})^2 \qquad \textbf{(D) }(2\pi)^2 \qquad \textbf{(E) }(3+\sqrt{2})(3-\sqrt{2}) \qquad$

Solution

After trying each option we have

A) $3^\frac{2018}{3}$ which is irrational as 2018 is not divisible by 3

B) $3^\frac{2019}{2}$ which is irrational as 2019 isn't divisible by 2

C) $3^2+\sqrt{2}^2+6\sqrt{2}$ which equals $11+6\sqrt{2}$ which is irrational

D) $(2\pi)^2$ equals $4\pi^2$, which is irrational

E) $(3-\sqrt{2})(3+\sqrt{2})=9-2=7$ which is rational

Our answer is $\boxed{\textbf{(E) }(3-\sqrt{2})(3+\sqrt{2})}$.

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