When the fraction 1/288 is expressed in base 12, is it terminating or repeating? Explain.

1/288 =0.003472222222 in base 10

0.003472222222222222 x12 =0.04166666 =0.0 in base 12

0.04166666 x 12 =0.5 in base 12=0.00 in base 12

0.5 x 12 =6 =0.006 base12. So, we have:

1/288 =0.006 in base 12 and it terminates.

Anyone? Please help.

That is really cluey guest.  I have not done a problem like this in yonks.

Here is another way to look at it.

$\frac{1}{288}\;base \;10 = \frac{1}{2*10^2+8*10+8}\;\;base 10$

But we want to try and convert this to base 12

288 is much bigger than 12 but how does it compare to 12^2

288=2*144=2*12^2

so

Since 288 is a multiple of a power of 12 then 1/288    MUST   be a terminating decimal in base 12

END OF QUESTION    

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So what it it???

$\frac{1}{288_{10}}=\frac{1}{2*12^2}\;(base10)\;\;=\frac{1}{2}of 0.01 \;\;(base\;12) = 0.006_{12} \quad\text{(because half of 12 is 6)}\\ or\\ 12^3=1728 \;\;and\;\; 1728\div 288=6\;\;\\so\\ \frac{1}{288_{10}}=\frac{1}{\frac{12^3}{6}}\;\;base10=\frac{6}{12^3}\;\;base10\;\;=0.006_{12}$

When the fraction 1/288 is expressed in base 12,

is it terminating or repeating? Explain.

$\begin{array}{|rcll|} \hline \frac{1}{288} &=& a\cdot 12^{-1} + b\cdot 12^{-2}+ c\cdot 12^{-3}+ d\cdot 12^{-4}+ e\cdot 12^{-5} \dots \quad & | \quad \cdot 12 \\ \frac{12}{288}=\frac{1}{24} &=& a + \underbrace{ b\cdot 12^{-1}+ c\cdot 12^{-2}+ d\cdot 12^{-3}+ e\cdot 12^{-4} \dots }_{\text{decimal places}=\frac{1}{24}}\\ integer(\frac{1}{24}) &=& a \quad \Rightarrow \quad a = 0\\\\ \frac{1}{24} -a &=& b\cdot 12^{-1}+ c\cdot 12^{-2}+ d\cdot 12^{-3}+ e\cdot 12^{-4} \dots \\ \frac{1}{24} -0 &=& b\cdot 12^{-1}+ c\cdot 12^{-2}+ d\cdot 12^{-3}+ e\cdot 12^{-4} \dots \\ \frac{1}{24} &=& b\cdot 12^{-1}+ c\cdot 12^{-2}+ d\cdot 12^{-3}+ e\cdot 12^{-4} \dots \quad & | \quad \cdot 12 \\ \frac{12}{24}=\frac{1}{2} &=& b + \underbrace{ c\cdot 12^{-1}+ d\cdot 12^{-2}+ e\cdot 12^{-3} \dots }_{\text{decimal places}=\frac{1}{2}}\\ integer(\frac{1}{2}) &=& b \quad \Rightarrow \quad b = 0\\\\ \frac{1}{2}-b &=& c\cdot 12^{-1}+ d\cdot 12^{-2}+ e\cdot 12^{-3} \dots \\ \frac{1}{2}-0 &=& c\cdot 12^{-1}+ d\cdot 12^{-2}+ e\cdot 12^{-3} \dots \\ \frac{1}{2} &=& c\cdot 12^{-1}+ d\cdot 12^{-2}+ e\cdot 12^{-3} \dots \quad & | \quad \cdot 12 \\ \frac{12}{2}= 6.0000000\dots &=& c + \underbrace{ d\cdot 12^{-1}+ e\cdot 12^{-2} \dots }_{\text{decimal places}=0}\\ integer(6.0000000\dots) &=& c \quad \Rightarrow \quad c = 6 \\\\ \frac{1}{288} = 0.006_{12}\\ \hline \end{array}$