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For how many integers between 1 and 11 (inclusive) is  n/12 a repeating decimal?

I can obviosly guess and check this, but I want the actual way. 

 Feb 22, 2019
 #1
avatar+4449 
+1

If a rational number's decimal representation doesn't terminate then it repeats.

 

A rational numbers decimal representation will terminate only if the denominator's only prime factors are 2 and 5.

 

With 12 in the denominator we need a 3 in the numerator to ensure the decimal terminates.

 

Thus 3/12, 6/12, 9/12 will all terminate

 

1/12, 2/12, 4/12, 5/12, 7/12, 8/12, 10/12, 11/12 will all repeat

 Feb 22, 2019
 #2
avatar+21848 
+3

 For how many integers n between 1 and 11 (inclusive) n/12 is a repeating decimal?

 

\(\begin{array}{|r|r|r|r|r|} \hline & & & & & \text{prime number only } \\ & & & \text{factorize} & \text{prime numbers} & 2 \text{ or } 5 \\ n & \dfrac{n}{12} & \text{cancel} & \text{the denominator} & \text{in the denominator}& \text{will terminate} \\ \hline 1 & \dfrac{1}{12} & \dfrac{1}{12} & \dfrac{1}{2^2\cdot 3} & 2,\ 3 \\ \hline 2 & \dfrac{2}{12} & \dfrac{1}{6} & \dfrac{1}{2 \cdot 3} & 2,\ 3 \\ \hline 3 & \dfrac{3}{12} & \dfrac{1}{4} & \dfrac{1}{2^2} & 2 & \checkmark \Rightarrow \text{terminate} \\ \hline 4 & \dfrac{4}{12} & \dfrac{1}{3} & \dfrac{1}{3} & 3 \\ \hline 5 & \dfrac{5}{12} & \dfrac{5}{12} & \dfrac{1}{2^2\cdot 3} & 2,\ 3 \\ \hline 6 & \dfrac{6}{12} & \dfrac{1}{2} & \dfrac{1}{2} & 2 & \checkmark \Rightarrow \text{terminate} \\ \hline 7 & \dfrac{7}{12} & \dfrac{7}{12} & \dfrac{7}{2^2\cdot 3} & 2,\ 3 \\ \hline 8 & \dfrac{8}{12} & \dfrac{2}{3} & \dfrac{2}{3} & 3 \\ \hline 9 & \dfrac{9}{12} & \dfrac{3}{4} & \dfrac{3}{2^2} & 2 & \checkmark \Rightarrow \text{terminate} \\ \hline 10 & \dfrac{10}{12} & \dfrac{5}{6} & \dfrac{5}{2\cdot 3} & 2,\ 3 \\ \hline 11 & \dfrac{11}{12} & \dfrac{11}{12} & \dfrac{11}{2^2\cdot 3} & 2,\ 3 \\ \hline \end{array} \)

 

see: https://gmatclub.com/forum/math-number-theory-88376.html

 

laugh

 Feb 22, 2019

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