## Re: maths questions

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• I believe (and I hope I m right) that if a two-digit number to the power of seven yields a number whose last two digits end in x and y (for example), then any
Message 1 of 4232 , Sep 17, 2001
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I believe (and I hope I'm right) that if a
two-digit number to the power of seven yields a number
whose last two digits end in x and y (for example),
then any number ending in those two digits will, when
raised to the power of 7, end in x and y. So if 34^7
ends in 44 (52523350144 to be precise), then 3234^7
should end in 44. And it does
(3699815853555305399228544). <br><br>In this way, all we have to do is raise 7
to the power of 7, raise the last two digits to the
pwer of seven, and repeat this till we get a
pattern.<br><br>Let s(x) mean the the last two digits of the number
got when raising 7 to the power of 7 x times. So...
<br>s(1) = last two digits of 7^7 = 43<br>s(2) = last two
digits of (7^7)^7 = last two digits of 43^7 = 07<br>Get
me?<br><br>s(1) = 43<br>s(2) = 07<br>s(3) = 43<br>s(4) =
07<br>etc etc etc. Very boring, ain't it? So for every even
x, s(x) = 07. Answer: s(1000) = 07
• can t sleep (yawn), i spent around 3 hr.s on your your original post here. it seems this implicit function solution set is the semi-stable modular eliptical
Message 4232 of 4232 , Apr 25, 2011
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can't sleep (yawn), i spent around 3 hr.s on your your original post here. it seems this implicit function solution set is the semi-stable modular eliptical curves A. Wiles gave in his proof of fermat's last theorem. Your function is way simpler. have to get some sleep before i got to go to work, but the proof of this is around 4 to 5 pages as you said. i'll have to look at it more this evening. also, what's the scoop of you being the source for the plot of tron legacy? we should all get royalties! heheh.
--- In mathforfun@yahoogroups.com, jeshields_98 wrote:
>
> I hit the wrong button...<br><br>It should be = c^n<br>not + c^n<br><br>very VEEERRRRRYYYYY important<br><br>sorry
>
I
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