+118690 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 
+118690 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
