Re: Power Tower and Carmichael Lambda Function

*To*: mathgroup at smc.vnet.net*Subject*: [mg89739] Re: [mg89662] Power Tower and Carmichael Lambda Function*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>*Date*: Thu, 19 Jun 2008 05:44:44 -0400 (EDT)*References*: <200806170436.AAA26560@smc.vnet.net>

I do not understand why you did not just do this: Fold[PowerMod[#2, #1, 10^4] &, 1, Range[2008]] // Timing {0.00546, 1008} It saves you having to use CarmichaelLambda, is not too long to wait, and leaves not doubt as to the correct answer. Andrzej Kozlowski On 17 Jun 2008, at 13:36, John Snyder wrote: > On New Year's Day of this year the following problem was published > in a > problem contest on an internet website asking for the last four > digits of > the following number: > > 2008^(2007^(2006^.^.^.^(2^1))) > > For some reason I got interested in the problem (it isn't homework, > I am 58 > years old) and I solved it using Mathematica's CarmichaelLambda > function as > follows: > > In[9]:= lambda=FoldList[CarmichaelLambda[#1]&,10000,Range[6]] > Out[9]= {10000,500,100,20,4,2,1} > > In[10]:= m=Range[2002,2008]//Reverse > Out[10]= {2008,2007,2006,2005,2004,2003,2002} > > In[11]:= Mod @@@ Transpose[{m, lambda}] > Out[11]= {2008,7,6,5,0,1,0} > > Now, since the tower of powers is evaluated from the top down I > reasoned > that everything from the top down through the 5 contributed nothing > to the > answer because 5^0=1. I got my answer doing the following: > > In[12]:= PowerMod[2008,7^6,10000] > Out[12]= 5328 > > When the solution was finally published the other day the following > result > was listed as being the correct solution: > > In[13]:= PowerMod[2008,7^6^5,10000] > Out[13]= 1008 > > It seems to me that the relevant exponent should be just 7^6, not > 7^6^5 > because 5^0=1. Playing around with some smaller and easier numbers in > Mathematica I think I am correct, but the problem site says > otherwise. I am > no expert in number theory so could someone who is please explain to > me > which answer is correct and why? Does Mathematica really use the > parentheses > in the exponent in evaluating something like this? > > Thanks, > > John

**References**:**Power Tower and Carmichael Lambda Function***From:*"John Snyder" <jsnyder@wi.rr.com>