. [Q-11]If X^Y denotes X raised to the power of Y, find out last two
digits of (2957^3661)+(3081^3643)
Options
o 42
o 38
o 98
o 22

• 98 is the ans
  7^1 is 7
  and
  5*1=5
  so last to digit of (2957^3661) is 57
  
  same for
  1^3 is 1
  8*3=24 unit digit of 24 is 4
  last 2 digit of(3081^3643) is 41
  41+57=98....
• 12 years agoHelpfull: Yes(46) No(7)
• my answer is comng out to b 98. pls reply if is correct..?
• 12 years agoHelpfull: Yes(14) No(1)
• 98 is correct
• 12 years agoHelpfull: Yes(6) No(0)
• i think 98
  
• 12 years agoHelpfull: Yes(5) No(2)
• 98 is correct.
• 12 years agoHelpfull: Yes(3) No(2)
• ya 98 nly...
  
• 12 years agoHelpfull: Yes(2) No(1)
• ans is 98...
• 12 years agoHelpfull: Yes(2) No(1)
• 12345.....424344 = (1234...424290) + (54)(hint:last 4 digits 4290+54=4344).
  first part we take 123456........424290(in this last digit should be 0 so it can divisible by 5,and also divisible by 9)so it can divisible by 45.
  second part 54 divide by 45 we get remainder 9 so answer is 9
• 12 years agoHelpfull: Yes(1) No(15)
• answer is 22.
• 12 years agoHelpfull: Yes(1) No(12)
• 8 will be last digit
  reason: 7^4 will give last digit 1 so 1st expression will be 2957^4^739 *2957=last digit will be 1*7
  similarly last digit of 2nd expression is 1 hence 7+1=8;
  for 2nd last digit 5+4=9
  
• 12 years agoHelpfull: Yes(1) No(5)
• sol:(2957)^(3661) => 7^1=7
  5*1=5
  =>last to num of this is 57
  same way 3081^3643 => 1^3=1
  8*3=24
  => 241
  sum of last to num => 41+57=98 :)
  
• 12 years agoHelpfull: Yes(1) No(0)