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Algebra Difficulty 6.1 National Olympiad Prove it Romania

We say that a simple periodic decimal fraction ff has the reduced length equal to nn (where nn is a positive integer) if ff has a nn-digit period and ff cannot be represented as a simple periodic decimal fraction with a period having less than nn digits. For instance, 0.(223)0.(223) has the reduced length 3, while 0.(2323)0.(2323) has the reduced length 2, as 0.(2323)=0.(23)0.(2323) = 0.(23).

a) Prove that f=0.(2)⋅0.(3)f = 0.(2) \cdot 0.(3) is a simple periodic fraction with reduced length 3.

b) Does there exist two simple periodic fractions with reduced length 1, such that their product has reduced length also 1?

c) Does there exist two simple periodic fractions with reduced length 3, such that their product has the reduced length also 3?

Solution

a) f=0.(2)⋅0.(3)=29⋅39=227=0.(074)f = 0.(2) \cdot 0.(3) = \frac{2}{9} \cdot \frac{3}{9} = \frac{2}{27} = 0.(074) is a fraction of reduced length 3.

b) Yes. For an example, 0.(3)⋅0.(6)=0.(2)0.(3) \cdot 0.(6) = 0.(2).

c) Yes. For example, 0.(270)⋅0.(370)=2709⋅3⋅37⋅370999=100999=0.(100)0.(270) \cdot 0.(370) = \frac{270}{9 \cdot 3 \cdot 37} \cdot \frac{370}{999} = \frac{100}{999} = 0.(100).

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