Convert the following rational number into decimal and write the periods and their length of 7/5
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Step-by-step explanation:
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
1
13
is 6 long. It will always be a factor of Euler's totient function of the denominator. For prime p, that is p−1
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