If we consider not base 10 but rather base 100, 1000, etc, then the requirement that n^3 have all odd digits using base 10^K is equivalent to every Kth digit being odd back in normal base 10. This is a different "test set" than Victor Miller was testing by asking that n^3 mod 10^k have all k digits odd; this asks for the least-sig k digits to be odd. Similarly by asking about 10*n^3 in base 10^K we could get a still different "test set" etc. --- Another thing one could consider is "carryless" problem, that is, consider not numbers but rather polynomials in a dummy variable with integer coefficients. Then ask about whether both P(x) and P(x)^3 can have all coeffs odd. This can also be done for formal power series instead of polynomials. Can we find such a series, all of whose infinite number of terms are odd, and its cube ditto? Answer=NO because (a0+a1*x+a2*x^2+a3*x^3+...)^3 = a0^3 +3*a0^2*a1*x +(2*a0^2*a2+a0*a1^2+2*a1^2*a0+a2*a0^2)*x^2 +... and the coefficient of x^2 is always even.