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![[Post New]](/templates/default/images/icon_minipost_new.gif) 18 May 2007 07:48:36 IST
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tnx arvind bro..! for the timely help..!!!
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B.Tech CSE
NIT Trichy |
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18 May 2007 10:06:03 IST
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i hav found the error in my ans. & so, i will edit my solution...
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PLEASE RATE MY ANSWERS IF YOU FIND THEM USEFUL... |
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18 May 2007 11:27:35 IST
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hey i found ANOTHER solution of this
see
(7)^256 can be written as
(289)^128
which can be written as
(280-1)^128
if we expand this term,we get
128C0 290^128+ 128C1 290^127 - 128C3 290^126....................
................................. 128C126 290^2- 128C1 290+1
RIGHT???
now the terms 10^3 can be taken common,
10^3(m) +128C126 290^2- 128C1 290+1
as it is 10^3 therefore the last three will be determined by the last three terms........
128C126 290^2- 128C1 290+1=683564800 -37120+1=
683527680+1=683527681
hence the last three digits are 681
and the expansion can be written as
10^3(n)+681
this method requires a trock of calculations which cm, through practice....
PLS RATE MY EFFORTS....
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IIT- Imposible Is This(atleast fr meeeeeeeee) |
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18 May 2007 20:21:35 IST
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what happened???/
is my method wrong ???
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IIT- Imposible Is This(atleast fr meeeeeeeee) |
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18 May 2007 23:39:18 IST
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not at all nick
u r right
infact i too approached the same way
bt may b becozz of the calculation mistake in the last step....... it went wrong.............
ur approach nd ans both r right
take my salute
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there is no right way 2 do something wrong !!!!!!!! |
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18 May 2007 23:54:06 IST
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any other method guys..!??
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B.Tech CSE
NIT Trichy |
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19 May 2007 07:52:27 IST
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ne1 plz....!?
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B.Tech CSE
NIT Trichy |
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19 May 2007 22:19:04 IST
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THESE MANY METHODS ARE NOT ENOUGH??!!!
PHEWWWWWWWW
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IIT- Imposible Is This(atleast fr meeeeeeeee) |
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