Question

Write the length and period of 7 upon 15

Answer 1

Answer:

Assuming there are no factors of 2,5 in the denominator, one way is just to raise 10 to powers modulo the denominator. When you find −1 you are halfway done. Taking your example: 102≡9,103≡−1,106≡1(mod13) so the repeat of 113 is 6 long. It will always be a factor of Euler's totient function of the denominator. For prime p, that is p−1.