Show that there are infinitely many primes p such that 1/p has its shortest period divisible by 3: 1/p = 0.a1a2 ... a3ka1a2... . Amongst such primes find the largest possible value for ai + ai+k + ai+2k. For example, 1/7 = 0.142857... So the possible values for p = 7 are 1+2+5 = 8 and 4+8+7 = 19.
Solution
The shortest period is the smallest d such that p divides 10d - 1. [We know, for example, that p divides 10p-1 - 1, so such d exists, but that line of thought does not lead anywhere.]
Let p be a prime and let Np = 102p + 10p + 1. Obviously Np = 3 mod 9, so Np is divisible by 3 but not 9. So take q to be a prime dividing Np/3. We have just established that q is not 3. Now 103p - 1 = Np(10p - 1), so q divides 103p - 1. Hence the shortest period d for q must divide 3p. So it must be 1, 3, p or 3p.
It cannot be 1 for then q would divide 10 - 1 = 9 and we have shown that q is not 3. It cannot be p, for then Np = 1 + 1 + 1 = 3 mod q, whereas we are assuming that q divides Np. If it is 3, then q divides 103 - 1 = 3337 and so q = 37. Now we claim that we can always choose q to be a prime other than 37. If this were not true then we would have Np = 3.37k. But then Np = 3 mod 4, whereas Np = 1 mod 4. So d must be 3p. Thus by letting p run through the primes, we get a sequence of primes q with shortest period 3p. Each prime q must be distinct, since each has a distinct shortest period. So we get infinitely many primes q with shortest period a multiple of 3.
Suppose 1/q = 0.a1a2a3 ... with shortest period 3p. Then bj = 0.ajaj+1aj+2 ... is the fractional part of 10j-1/q. Obviously aj < 10 bj, so aj + aj+p + aj+2p < 10(bj + bj+p + bj+2p) = (1/q) (10j + 10j+p + 10j+2p = 10jNp/q. But Np/q is an integer, so 10(bj + bj+p + bj+2p) is a multiple of 10. Each bi < 1, so 10(bj + bj+p + bj+2p) ≤ 20 and hence aj + aj+p + aj+2p ≤ 19. But 19 can be achieved, as illustrated by the case 1/7.
© John Scholes
jscholes@kalva.demon.co.uk
10 Oct 2002