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Let $R$ be the set of primitive $42^{\text{nd}}$ roots of unity, and let $S$ be the set of primitive $70^{\text{th}}$ roots of unity. How many elements do $R$ and $S$ have in common?

 Jan 17, 2019

Best Answer 

 #1
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\(GCF(42,70)=7\\ \text{Thus the common roots of unity are}\\ \Large e^{\pm i 2\pi \frac{k}{7}},~\normalsize k=0,1,\dots ,6\)

 

\(\text{There are 14 of these}\)

.
 Jan 17, 2019
 #1
avatar+4810 
+2
Best Answer

\(GCF(42,70)=7\\ \text{Thus the common roots of unity are}\\ \Large e^{\pm i 2\pi \frac{k}{7}},~\normalsize k=0,1,\dots ,6\)

 

\(\text{There are 14 of these}\)

Rom Jan 17, 2019

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