Date: Dec 6, 2012 6:38 PM
Author: Dr J R Stockton
Subject: Re: No Putnam spoilers please
In sci.math message <Pine.NEB.4.64.1212041922190.1746@panix1.panix.com>,

Tue, 4 Dec 2012 19:48:01, William Elliot <marsh@panix.com> posted:

>On Tue, 4 Dec 2012, Dr J R Stockton wrote:

>> Sun, 2 Dec 2012 19:47:28, William Elliot <marsh@panix.com> posted:

>> >On Sat, 1 Dec 2012, William Elliot wrote:

>>

>> >> "What's 5^2012 mod 7?" is a Putnam question? ;-}

>

>> >phi(7) = 6; 5^2012 = 5^(6 * 335 + 2) = 5^2 = 25 = 4 (mod 7)

>>

>> Correct by direct calculation in Bases 7, 10, 13, using my longcalc.exe.

>

>For all n in N, 5^n = 5 (mod 10)

>Proof by induction. 5^1 = 5 (mod 10).

>If 5^n = 5 (mod 10), then 5^(n+1) = 5^n * 5 = 5 * 5 = 25 = 5 (mod 10)

>

>phi(13) = 12;

>5^2012 = 5^(12 * 168 + 6) = 5^6 = 25^3 = (-1)^3 = -1 = 12 (mod 13)

>

>1200

> 720

> 96

>

>longcalc.exe is too smart for it's britches.

In that calculation, longcalc uses only elementary arithmetic, as used

to be taught in schools in my day. Your proof, however, is 50%

incomprehensible to me. Though it may well be right. Longcalc found an

error in *a* printed representation of (3^349-1)/2, which you might have

difficulty with. Fx : checks : I think it has found another one.

--

(c) John Stockton, Surrey, UK. ?@merlyn.demon.co.uk Turnpike v6.05 MIME.

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