Let S[x] be the digital root function (also known as the repeated digital sum function), where one adds the digits of positive integer x, then adds the digits of the sum until obtaining a single-digit number. (For example, S[975] = 3 because 9 + 7 + 5 = 21 and 2 + 1 = 3).

Given S[a^a] = 2, what is the smallest positive integer that a can be such that a is a perfect power?


Note: a is a perfect power if there exist natural numbers m > 1, and k > 1 such that m^k = a.

(In reply to re: Extra Credit (spoiler) by Dej Mar)

log(5) = 0.698970004
log(78125) = log(5^7) = 7*log(5) = 4.89279028,
       so 78125 has 5 digits
log(78125^78125) = 78125*log(78125) = 382249.24,
       so 78125^78125 has 382,250 digits


log(11) = 1.04139269
log(161051) = log(11^5) = 5*log(11) = 5.20696345,
       so 161051 has 6 digits
log(161051^161051) = 161051*log(161051) = 838586.67,
       so 161051^161051 has 838,587 digits

But that's pretty much the limit of accuracy on my calculator.  No extra credit for me.  ;-(

Posted by Steve Herman on 2008-07-14 10:33:59