Magic prime fraction period

Computer Science Level 4

For a prime number p 1 = 3 p_1 = 3 and for square p 1 2 = 9 p_{1}^2 = 9 , the length of the period of the fraction L ( 1 3 ) = L ( 1 9 ) = 1 L(\frac{1}{3})=L(\frac{1}{9})=1 .

Find another minimal prime number p 2 > 3 p_2>3 for which L ( 1 p 2 ) = L ( 1 p 2 2 ) L(\frac{1}{p_2})=L(\frac{1}{p_{2}^2}) .

Give answer p 2 p_2 .

We exclude prime number equal 2 and prime number equal 5 that the period length there is equal to zero.

Problem from "Kvant" - Soviet and Russian popular scientific physics and mathematics journal for schoolchildren and students. There is the sequence A002371 .

Bonus. Open question - How many such numbers exist? It is very difficult to find p 3 , p 4 , . . . p_3, p_4, ... - for example p 3 > 56000000 p_3>56 000 000 .


The answer is 487.

Yuriy Kazakov by Yuriy Kazakov , 61, Russia

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2 solutions

Yuriy Kazakov
Jan 10, 2021

I also use python for solution. And I try to find p 3 p_3 - but it very big time. Here A045616 there is the result for p 3 p_3 .

1 Helpful 0 Interesting 0 Brilliant 0 Confused
K T
Jan 9, 2021

output:

per(1/2)=0; per(1/4)=0

per(1/3)=1; per(1/9)=1

per(1/5)=0; per(1/25)=0

per(1/487)=486; per(1/237169)=486 

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def FillPrimes(upto):
    primes=[2]
    for n in range(3, upto):
        for p in primes:
            if p**2>n:
                primes.append(n)
                break
            if n%p==0:
                break
    return primes

# do a long division, keep track of repetitons 
def FractionExpansionPeriod(numer, denom, base, cutoff=0):
    quot = numer // denom
    numer = (numer - denom * quot) * base
    numers=[]
    while not (numer in numers):
        if numer==0:
            return 0
        if cutoff!=0 and len(numers)==cutoff:
            return -cutoff
        numers.append(numer)
        digit=numer//denom
        numer = (numer-denom*digit)*base
    return len(numers)

for p in FillPrimes(1000):
    per1=FractionExpansionPeriod(1,p,10)
    per2=FractionExpansionPeriod(1,p*p,10, per1)
    if per1==per2:
        print ("per(1/{})={}; per(1/{})={} ".format(p,per1,p*p,per2))

0 Helpful 0 Interesting 1 Brilliant 0 Confused

Thanks for attention.

Yuriy Kazakov - 5 months ago

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