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I am looking for someone to help talk me through some Abstract homework.
-integers and modulo n problems. Euler's function...
1. compute the remainder when 37^100 is divide by 29.
2. compute the last 2 digits of 9^1500
Last edited by parthenos (2008-08-25 12:27:08)
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^ exponent
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1. The key here is of course everything (as you seem to know) is modulo 29. What can you immediately reduce the problem to?
2. Same as above, but you would be calculating mod 100.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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I cant make head or tail of Rickys hint, so I doubt if Parthenos can either. Here are my own hints instead.
#1
By Fermats little theorem, if x is not divisible by 29,
Now
. etc. Continue this way until .#2
Keep taking powers of 9 until you get to a number whose last two digits are 01.
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I'm pretty sure Ricky's hint was just saying that 37 could be changed into 8.
Why did the vector cross the road?
It wanted to be normal.
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Okay, maybe you dont need to calculate
directly in the second one. You can stop at .Notice that
, where N is an integer.Offline
Thanks for you help Jane. I still don't quite understand how I find the last 2 digits in #2.
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To get the last 2 digits you need the remainder modulo 100. First find the phi of 100:
And if you don't beleive, the number is actually
2310809578111909272693109431184832846484968454396283812529115413319435556973292122101720139716262409469889717513948376726158501486182636383531313869623735751595188198743086350673413093292498741784795660389148326466211372843337723144590341000538769498117226562808444716579845071408688720747381120135479199680471885180482121554984483377022066232113498842614313541610752653690490275184816645775562870080222550532658199952189433551842563700110684899935346509404097452154895288699360903786484615575470777362920177718027703180305669825349020152868844727956235156059960772077214312800556301983539901820176796454299860300131486822074451633519214994291242241850066416567586776526981799888092096865174444248800546127994730384671200620078154936387315118122040519117349396356198197315367766646921156257797160661163192060277301722871797101013527586327674333920080776435765228230385762165404932957246335621254520607306174400378047735376342518713628466946614321497738427647167939078993913147690702992638955965837451910388196619417991842556404018337740923222175380153877003983519741457506625666074023573361964063311318755920947209571433341645962479076131201964416276406387762072361807631460445372486964777059883706070699922193176468574966898852772974365576953262531708962699524486008324366541931862849776343129934761158587798436540362085500749517170402734545433993099089688531543345393342814833105525138762128016370025483881594201769798453765660001
IPBLE: Increasing Performance By Lowering Expectations.
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4|150
It does???!!!
Last edited by JaneFairfax (2008-08-26 11:08:32)
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krassi_holmz wrote:4|150
It does???!!!
You got me!
So... Since 1500 = 37.40 + 20, then
Last edited by krassi_holmz (2008-08-26 10:53:19)
IPBLE: Increasing Performance By Lowering Expectations.
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