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Define the sequence of positive integers a_n recursively by a_1=7 and a_n=7^(a_n-1) for alln> =2. Determine the last two digits of a_2007.

 

Find the last 7 digits in the binary expansion of 27^1986. (Give your answer in binary.)

thanks!

MIRB16  Aug 27, 2018
 #1
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+3

27^1986 =  48801775164233 .............493428425689 (base 10)=100001111011001(base 2)

Guest Aug 27, 2018
 #2
avatar+7025 
+3

27^1986 = 48801775164233180193152186583304417336797680575180581115895393460515804997481426460328920628366330784510548613573893133597581628998231738095941136304757747700966133994279657641604503248144993063886307509474264646012144202451717134988223435658134723146939548526526046746149774508324183700407959892199488896439369356604901578709343457074224182222533446701001327664448924429135151145101780007678647192526652283864386426058549923535427662270800539978182246623630215550544972216345285441276619530080398735910161756589449840397939641361894901562665980047466740082396373341778906897800555115722452770664688076387259535876436046972899204638534393769005219725810593392077925096955164314720390198579103842847954992632961329117315094216615129428536324654588082306372021647022430577611704382369124740553707849232982840968333108506219759397531460279020717574204357008717885397405312628389031757241409912778322513652225905208643086567820940013264513686836151697257293488310351014146560735095573779572404778048653597239139290808021905056342235736193779153753316842586904178196600043336256641895973206688460099090120406045255624825627683977323696929363287506689433912930661175286684677160661203889391570143445392987720450701894381239294874928803004177216894164745421763945685814401999598340303489036530867374232572727166897181905438211435777103860575566122738459632273930151354490629643457508872730539666645848055176086262500173450757837780931124099511147443471470828134586813514044949515967779408501414682996812976558402552553142807980868977183835493023048436656516057263092336591331513716603390064194308624685089075088054532483604955744721552650388426853236066590978184087032928185890064270903768451944629764513609533788042180992515318842081697688028887938906492323565305085808157568546090292860937463322677394902590580918646428717626492549592383518428731724982544318398065913196769349927074541649454968742704298225424732774113924319005257485049325235138238093730273646224858897535235298885105594129943937351459724702482583667236536502693815709185789808126712796159398263200073561808422405408703688494449817834260263098757025774244181566762357315704174052124364390755529262395531978098406249555755014544306699077239975625672904487308441554678704578185924139361375521658379381070765883331271855520007826465075200564387537336981534406153890434416074062845454707273089136109080238153308852346576731949235793839884975531153052962588362467246851989966711345919740451341621252450778752838477266614739871237844364433820224901541153831413862665959090015057100579979531958425068248597012156787283346913344245335741283554176163827879328047740051313782996843172666708463050608416638964963699210219313787074584062514104031192818402598712368534388109017171715451967516936812146050223558204491606779630732845610748256709841526809731630661112279428213058346512493428425689

27^1986 = 89 mod 128

Therefore last 7 digits = 1011001.

I calculated it using a program, so... :P

MaxWong  Aug 27, 2018
 #3
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0

Happy birthday max!

Guest Aug 27, 2018
 #4
avatar+93920 
+1

Is it really your birthday Max?      frown

 

 

 

Happy Birthday !!   

 

生日快樂              laugh

 

Alles Gutes zum Geburtstag !!

Melody  Aug 27, 2018
 #5
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0

I think the sequence goes like this:
7, 7^0, 7^1, 7^2, 7^3, 7^4, 7^5, 7^6, 7^7, 7^8...........etc. So the pattern for a(n) appears to be: 7^(n - 2). Therefore, a(2007) =7^(2007 - 2) =7^2005.
7^2005 =  263 979 516 .................................... 088 416 807        

Guest Aug 27, 2018
 #6
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+1

This is not what the sequence is.

 

a1=7, and for all n>=2 an=7an-1,

 

 

so a2=7a2-1=7a1=77=823543

Guest Aug 27, 2018
 #7
avatar+738 
+1

THANK YOU ALL AND HAPPY B-DAY MAX

MIRB16  Aug 28, 2018

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