For prime lovers

[Abdul Muntakim Rafi]

What are the primes p > 0 for which $1/p$ has a purely periodic decimal expansion with a period$5$ digits long?

[NOTE: $1/37 = 0.027027027027 .....$ starts to repeat immediately, so it is purely periodic. Its period is $3$ digits long.]

Hope u enjoy solving the problem...

[Abdul Muntakim Rafi]

Did anyone solve it or make any progress?

[mahathir]

The period of the repeating decimal of $1/p$ is equal to the order of $10$ modulo $p$. If $10$ is a primitive root modulo $p$, the period is equal to $p − 1$; if not, the period is a factor of $p − 1$. This result can be deduced from Fermat's little theorem, which states that \[10^{p-1}\equiv 1(mod p)\]\

So,all thats left is to find the prime numbers.I hope it was helpful.

[Abdul Muntakim Rafi]

Mahathir, how can you find the answer easily this way?

There is an easier way to find out the primes... You just need a crux move...

Here's it.

$1/P = 0.abcdeabcdeabcde...................... $

What's $100000(1/P) - (1/P) $