From: Victor Miller <victorsmiller@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] n and n^3 have all digits odd Message-ID: <CAEe1e0zHEW=rufvv1R7MmW+kGY-54Und+MJhsuoZF+NuzBxw7w@mail.gmail.com> Content-Type: text/plain; charset=ISO-8859-1

I'm doing some calculations pertaining to this. For all positive integers k define the set

S_k := { 1 <= n < 10^k | OddDigits(n) & OddDigits((n^3) mod 10^k) }

It looks like (log # S_k)/k --> 1 (where log is the natural log). Notice that 10^k-1 is always in S_k.

--that "e" should come up here seems quite amazing. But anyhow, let me make a simple remark. Let T_k = {1 <= n < 10^k | OddDigits(n) } and obviously #T_k = 5^k. So assuming MIller correct, (#S_k)/(#T_k) behaves for large k approximately like (e/5)^k where e=2.71828... So now, the chance that all 3k digits of n^3 (it is about 3k digits long) all are odd, is presumably (e/5)^(3k). Hence, the total cardinality of the original problem's set R_k = {1 <= n < 10^k | OddDigits(n) & OddDigits(n^3)} is presumably about #R_k =approx= 5^k * (e/5)^(3k) = A^k where A = e^3 / 5^2 = 0.8034. This compares with Charles Greathouse's previously-posted estimate #R_k =approx= (5/8)^k where 5/8 = 0.625. Either way, the conclusion is the same -- the total cardinality of R_1 U R_2 U R_3 U ... is presumably finite - but it's now less clear than it appeared from Greathouse's original estimate.