# Thread: Calculating the period of a decimal

1. ## Calculating the period of a decimal

Find the delay and period of $\frac{60}{336}$.

How I started:

$\frac{60}{336}=\frac{5}{2^2 \cdot 7}$.

Hence the delay $\mu=\max \{0,2 \}=2$.

I've run into a bit of a problem when I try and find the delay.

I need to find n s.t $10^n=1 \bmod 7$.

Comparing this to Euler's theorem $\Rightarrow n= \phi(7)=6\Rightarrow \nu=6.$.

My mark scheme also agrees with this result. However, can it always be done in this way? ie. will the delay always be reduced to $10^n=1 \bmod p$ where p is prime?

2. you are talking about period..right.
Please check your post...it is confusing!!!!

3. The answer to your second question is not quite. To find the repeating decimal for 1/q where q has no factors of 2 or 5 you need to find 10^n mod q = 1 and the answer will be the lowest common multiple of the n's that solve 10^n mod p^a = 1 where p is prime and a is the highest power of the prime that divides q.
For example 1/27 has period 3 even though 1/3 has period 1: you really have to solve 10^n mod 27 =1
1/91 has period 6 because both 1/7 and 1/13 have period 6.
But 1/119 has period 48 because 119=17*7 and 17 has period 16 and 7 has period 6

Euler's totient function has nothing to do with the period of a decimal, so the fact that 7 has period 6 and phi(7)=6 must be a coincidence. The definition of the totient function has nothing to do with our use of 10 as our number base. 1/7 has period 1 in base 7, period 3 in base 4.