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find the base of numeration in which 135/21 has a zero remainder

Guest Feb 7, 2018

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 #1
avatar+91795 
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find the base of numeration in which 135/21 has a zero remainder

 

Let the base be x

\(\frac{x^2+3x+5}{2x+1}=n \qquad n\in Z\\ \)

 

I am going to try trial and error.

x has to be 6 or bigger because of the 5.

 

If x=6 this becomes

\(\frac{36+16+5}{13}=\frac{57}{13}\ne n\)

 

If x=7

\(\frac{49+21+5}{14+1}=\frac{75}{15}=5=n\)

 

So base 7 works     cool

Melody  Feb 7, 2018
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1+0 Answers

 #1
avatar+91795 
+2
Best Answer

find the base of numeration in which 135/21 has a zero remainder

 

Let the base be x

\(\frac{x^2+3x+5}{2x+1}=n \qquad n\in Z\\ \)

 

I am going to try trial and error.

x has to be 6 or bigger because of the 5.

 

If x=6 this becomes

\(\frac{36+16+5}{13}=\frac{57}{13}\ne n\)

 

If x=7

\(\frac{49+21+5}{14+1}=\frac{75}{15}=5=n\)

 

So base 7 works     cool

Melody  Feb 7, 2018

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