A sequence {S(x)}, each of whose terms is a positive integer, is defined as:

S(x) = x/2, if x is even, and:

S(x) = x+3, if x is odd

Given that n is an odd positive integer and, S(S(S(n))) = 27, determine sod(n).

Note: sod(n) denotes the sum of the base-10 digits of n.

(In reply to True but unprovable by Steve Herman)

another cycle is 3,6,3

s(3)=3+3=6

s(6)=6/3=3

s(3)=3+3=6

And this can be proven easily for all integers n.

first prove for n=1

s(1)=4

s(4)=2

s(2)=1

thus it cycles.

Now assume it cycles with either 4,2,1 or 3,6 for all integers less than n

if n is even the we have n=2k

s(n)=k<n and thus by assumption will eventually cycle

on the other hand if n is odd then n=2k+1

s(n)=2k+4

s(2k+4)=k+2

k+2<2k when k>2 and since 2 is part of one of the cycles then n will eventually cycle for all integer values.

Edited on March 2, 2009, 8:06 pm

Posted by Daniel on 2009-03-02 19:34:39