Revision as of 20:12, 8 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) $(2pi)^2$ equals $4pi^2$ , which is irrational $\[\]$ E) $(3-\sqrt{2})(3+\sqrt{2})=9-2=7$ which is rational We have $\boxed{\bold{E}}$ $(3-\sqrt{2})(3+\sqrt{2})$

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+=Problem 3=
+Which of these numbers is a rational number?
+<math>\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</math>
 ==Solution==
 After trying each option we have

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

Revision as of 20:12, 8 October 2025

Problem 3

Solution