Assume X is a positive integer. If you divide 1/X, you will get a number that eventually becomes periodic: 1/9= 0.111..., 1/4= 0.25000..., and so on. Let's call numbers like 1/9 "pure" periodic, since the fractional part is formed just by the periodic part.
Prove that:
1. For all X, you will get a periodic part, and its length will be less than X.
2. If X is even, 1/X cannot be "pure". What happens if X is odd?
3. For some X, 1/X is "pure", the period length is even, and you can split the period in two halves that sum up to all nines. For example, 1/7=0.142857 142857... and 142+857=999. Which are these X values?
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Submitted by Federico Kereki
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Rating: 3.0000 (1 votes)
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Solution:
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1. Divide 1 by N using the classic long division algorithm. If you ever get a zero remainder, from then on you will always keep getting zero as a remainder, so you got a period of length 1.
Otherwise, you can get at most N-1 different remainders, and on the N step (if not earlier) you must have repeated a previous remainder, and from that point onwards you will enter a cycle, of length less than N, as required.
2. If 1/N=0.PPP... and P is D digits long, 1/N= P(10^-D+10^-2D+...)= P.10^-D/(1-10^-D)= P/(10^D-1), so (10^D-1)/N=P. Since 10^D-1 ends in 9, it follows that if N is even (or multiple of 5), 10^D-1 cannot be a multiple of N, and thus cannot be "pure" periodic. For any N, its period length is the smallest D such that 10^D-1 is a multiple of N.
3. If 1/N=0.PQPQPQ... and P and Q are both D digits long, then 10^D/N= P.QPQPQP... and (10^D+1)/N=P+1 so we need that both 10^2D-1
and 10^D+1 must be multiples of N. Of course, if 10^D+1 is a multiple of N, since 10^D+1)(10^D-1)= 10^2D-1, then it suffices to require that 10^D+1 be a multiple of N for odd D, and that for no smaller D, 10^2D-1 be a multiple of N. |
