find the last two digits of 583^512?

• 61..
  considering last two digits
  83^1=83
  83^2=89
  83^3=87
  .....
  83^12=61
  .....
  83^19=47
  83^20=01
  83^21=83..
  therefore we see power repeats itself after 20.. so dividing 512 by 20, we get remainder=12 so 83^12=61 last two digits..
  
• 10 years agoHelpfull: Yes(21) No(4)
• 583^512
  taking last two digit and squaring i.e (83^2)^256 = 6889...so now we have to calculate 89^256
  similarly repeating the process as (89^2)^128 = 7921...now 21^128
  (21^2)^64 = 441^64...so 41^64
  (41^2)^32 = 1681^32...SO 81^32
  (81^2)^16 =6561^16...SO 61^16
  (61^2)^8 =3721^8....SO 21^8
  (21^2)^4 =441^4...SO 41^4
  (41^2)^2 =1681^2...SO 81^2
  81^2 = 6561 SO 61 IS IN THE LAST
• 10 years agoHelpfull: Yes(9) No(1)
• given:( 583^512)
  =( 583^4*128)
  =(_21^128) (since , its hard to explain here that how it comes , better you follow the link and try to learn the concept, http://totalgadha.com/mod/forum/discuss.php?d=4405 )
  = 61
  hence, 61 is the ryt ans
• 10 years agoHelpfull: Yes(8) No(0)
• 61 will be the answer
  
• 10 years agoHelpfull: Yes(7) No(0)
• given:( 583^512)
  =( 583^4*128)
  =(_21^128)
  = 61
  hence, 61 is possible.
• 10 years agoHelpfull: Yes(4) No(0)
• 61 is the ans....
• 10 years agoHelpfull: Yes(2) No(0)
• @SAMANVITHA I did it in a manual way. 583^16*32=(583^4)128 Here I did square of 83 and got 6889 but I considered only 89..Then I multiplied it with another 89.Hence got the last 2 digits of 583^4 which will be 21.After that the equation will be reduced to (21^4)32..We will get the solution by solving it in the previous method.
• 10 years agoHelpfull: Yes(2) No(2)
• ans will be 61.
• 10 years agoHelpfull: Yes(1) No(0)
• Yes 61 is the answer can u plz explain it @MITA PAN
• 10 years agoHelpfull: Yes(1) No(0)
• WILL IT BE 69?
• 10 years agoHelpfull: Yes(1) No(9)
• find 83^12 we wil get 61 as last 2 digits.. i think 61 will be the answer....
• 10 years agoHelpfull: Yes(1) No(0)
• 3^2=9
  8*2=16
  answer will be 69
  
  
  
• 10 years agoHelpfull: Yes(1) No(0)
• 49 are the last two digits.
• 10 years agoHelpfull: Yes(0) No(6)
• 49 are the last digits..guys its very simple
• 10 years agoHelpfull: Yes(0) No(10)
• 83^512=((83)^2)^255=(89)^255=(90-1)^255=(-1)^55+55*90(-1)^54(binomial expansion)=(-1)+55*90)=4950-1=4949..So,the last 2 digits=49
• 10 years agoHelpfull: Yes(0) No(4)
• 96 after real multiplication
• 10 years agoHelpfull: Yes(0) No(2)
• 61 are the last two digits.
• 10 years agoHelpfull: Yes(0) No(0)