find out last two digits of (2957^3661)+(3081^3643)
Options
o 42
o 38
o 98
o 22

• haha i got the answer its 98 see i read this link http://www.totalgadha.com/mod/forum/discuss.php?d=4405
  its like for the first number -------
  57^3661=(57^4)^915*(57)=(49*49)^915*(57)=(2401)^915*9(57)
  take tens digit which is 01 and then according to the trick from the link
  =10*unit digit of power=01^anuthing is same and 01*57=57
  second number 81^3643=8*3=24 so it is 41 and so adding 41+57=98
• 12 years agoHelpfull: Yes(21) No(3)
• yeah shyamlesh.... sorry every 1 logic z same but a mistake in calc...
  
  2957)^1=...57 as last two digts
  (2957)^2=...49
  (2957)^3=...93
  (2957)^4=...01
  (2957)^5=...57
  
  
  as we observed dat last digit repeats for every power of five...
  now consider (2957)^3661=(2957)^{3660) * 2957
  since 3660 is multiple of 4 it gives 01 as last two digits and again multiplying with 2957 we get 57 as last digits..
  
  similarly do for (3081)^3643... we get 41 as last two dits
  
  now finally adding both 57+41= 98 therefore 98 is last two digits...
• 12 years agoHelpfull: Yes(19) No(1)
• answer is 38..
  observe carefully..
  (2957)^1=...57 as last two digts
  (2957)^2=...49
  (2957)^3=...93
  (2957)^4=...61
  (2957)^5=...77
  (2957)^6=...49
  
  as we observed dat last digit repeats for every power of five...
  now consider (2957)^3661=(2957)^{3660) * 2957
  since 3660 is multiple of 4 it gives 61 as last two digits and again multiplying with 2957 we get 77 as last digits..
  
  similarly do for (3081)^3643... we get 61 as last two dits
  
  now finally adding both 77+61= 138 therefore 38 is last two digits....
• 12 years agoHelpfull: Yes(9) No(13)
• see it can also be solved as mentioned by nagraj for the frst term it is
  57^1=57 ^2=49 ^3=93 ^4=01 and ^5=57 so it follows a cycle of 4 so 57*915+1=57 and so for second term is 47 which sums up to 98 :DDD
• 12 years agoHelpfull: Yes(4) No(1)
• sorry its 98 wrong calculation
• 12 years agoHelpfull: Yes(2) No(0)
• ans is 42 i am sure about it
• 12 years agoHelpfull: Yes(0) No(12)