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Question

For any natural number n>1, write the infinite decimal expansion of 1n (for example, we write 12=0.49 as its infinite decimal expansion, not 0.5). Determine the length of the non-periodic part of the (infinite) decimal expansion of 1n.

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Solution

For any prime p, let vp(n) be the maximum power of p dividing n; i.e p1/2p(n) divides n but not higher power. Let r be the length of the non-periodic part of the infinite decimal expansion of 1/n.
Write
1n=0.a1a2.....ar¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯b1b2......b2.
We show that r = max(v2(n);v5(n)).
Let a and b be the numbers a1a2........arandb=b1b2....bs respectively.(Here a1andb1 can be both 0.) Then
1n=110r(a+k1b(10s)k)=110r(a+110s1)
Thus we get 10r(10s1)=n(10s1)a+b. It shows that rmax(v2(n);v5(n)). Suppose r>max(v2(n);v5(n)). Then 10 divides b-a.Hence the last digits of a and b are equal: ar=bs. This means
1n=0:a1a2.......ar1¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯b2b1b2.....bs1.
This contradicts the definition of r. Therefore r=max(v2(n);v5(n)).

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