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Replies: 15   Last Post: Dec 14, 2012 6:05 AM

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 William Elliot Posts: 2,637 Registered: 1/8/12
Posted: Dec 4, 2012 10:48 PM

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.

Date Subject Author
12/1/12 Dmitry Yu. Mitin
12/1/12 William Elliot
12/1/12 Bart Goddard
12/1/12 William Elliot
12/2/12 William Elliot
12/3/12 Bart Goddard
12/4/12 Dr J R Stockton
12/4/12 William Elliot
12/5/12 Butch Malahide
12/6/12 Dr J R Stockton
12/8/12 Dr J R Stockton
12/8/12 Richard Tobin
12/9/12 Wasell
12/13/12 Dr J R Stockton
12/14/12 Pubkeybreaker
12/5/12 Phil Carmody