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

(→‎Problem 3)
 
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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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