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Find the smallest positve $n$ such that:

\begin{align*}
N &\equiv 6 \pmod{12}, \\
N &\equiv 6 \pmod{18}, \\
N &\equiv 6 \pmod{24}, \\
N &\equiv 6 \pmod{30}, \\
N &\equiv 6 \pmod{60}.
\end{align*}

 May 11, 2018

Best Answer 

 #3
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+1

Melody:

The smallest N is 6. The LCM of[12, 18, 24, 30, 60] = 360. Therefore:

N = 360k + 6, where k = 0, 1, 2, 3.........etc.

 May 11, 2018
 #1
avatar+982 
+1

Hey Guest!

 

\(\begin{align*} N &\equiv 6 \pmod{12}, \\ N &\equiv 6 \pmod{18}, \\ N &\equiv 6 \pmod{24}, \\ N &\equiv 6 \pmod{30}, \\ N &\equiv 6 \pmod{60}. \end{align*}\)

 

LCM [12, 18, 24, 30, 60] = 360

 

N = 360k + 6. 

 

6 is your answer.

 

Guest is right, my previous answer was wrong. 

 

I hope this helped,

 

gavin

 May 11, 2018
edited by GYanggg  May 12, 2018
 #2
avatar+101725 
+1

 

 

The smallest N is 6, the rest of my answer was incorrect. 

 May 11, 2018
edited by Melody  May 12, 2018
 #3
avatar
+1
Best Answer

Melody:

The smallest N is 6. The LCM of[12, 18, 24, 30, 60] = 360. Therefore:

N = 360k + 6, where k = 0, 1, 2, 3.........etc.

Guest May 11, 2018
 #4
avatar+101725 
0

Yes you are right .... 

I appologise for my incorrect answer :)

Melody  May 12, 2018

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