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ILL 1986 Problem 67

    Let f(x)=xnf(x) = x^n where nn is a fixed positive integer and x=1,2,⋯ .x =1, 2, \cdots . Is the decimal expansion a=0.f(1)f(2)f(3)...a = 0.f (1)f(2)f(3) . . . rational for any value of nn?

    The decimal expansion of a is defined as follows: If f(x)=d1(x)d2(x)⋯dr(x)(x)f(x) = d_1(x)d_2(x) \cdots d_{r(x)}(x) is the decimal expansion of f(x)f(x), then a=0.1d1(2)d2(2)⋯dr(2)(2)d1(3)...dr(3)(3)d1(4)⋯ .a = 0.1d_1(2)d_2(2) \cdots d_{r(2)}(2)d_1(3) . . . d_{r(3)}(3)d_1(4) \cdots .

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