If X^y denotes x raised to the power y, then find last two digit of
1507^3381+1457^3757

• 1507^3381+1457^3757
  => 07^3381+57^3757
  49^1690×7 + 49^1878 × 57
  => (100-51)^1690 ×7 + ( 100-51)^1878×57
  => 51^1690 × 7 + 51^1878×57
  > 07+57=64
• 8 years agoHelpfull: Yes(13) No(3)
• 07^3381 = 343^1127 = 43^1127 = ( 43^1124 )*( 43^3 ) = 01*07 = 07
  57^3757 = ( 57^3756 ) * 57 = ( 3249^1878 )*57 = ( 49^1878 )*57 = ( 2401^939 )*57 = 01*57 = 57
  so last two digit will be = 07+57 = 64
• 8 years agoHelpfull: Yes(6) No(8)
• 1507^3381+1457^3757
  for 1507^3381 keep the unit digit as it is &multply the second digit bypower of the first digit.Hence
  it becomes 07 and similarly second term of the given problem becomes 57
  then 07+57=64
  
• 8 years agoHelpfull: Yes(2) No(2)
• 1507^7 + 1457^3737
  we check the last two digits.
  hence 07^1 + 57^7
  hence 07 + 93
  hence answer = 00
• 8 years agoHelpfull: Yes(1) No(11)
• 43^1124=...01 how we get to know?
• 8 years agoHelpfull: Yes(0) No(0)
• 1507*1507^3380 +1457*1457^3756
  1507*49^1690 +1457*49^1878
  1507*51^1690 +1457*51^1878
  1507*01 +1457*01
  1507 +1457
  07 +57=64
• 8 years agoHelpfull: Yes(0) No(2)
• 07^4=01,17^4=21,27^4=41,37^4=61,47^4=81 after that again increases from 01 to 81
  so 57^4=01
  so 1457^3757=57^3757=(57^4)^939*1457=(01)^939*57=57
  so last two digit 64
  
• 8 years agoHelpfull: Yes(0) No(2)
• 7^1=7 , 7^7=49
  7+9=16
• 8 years agoHelpfull: Yes(0) No(3)
• 07^1+57^1
  07+57
  64
• 8 years agoHelpfull: Yes(0) No(3)
• 7^1+7^7=7+0=7 unit digit
  for 10th place (10 place digit in base*unit digit in power)=0*1+5*7=0+35
  add separately unit digit to last digit of the 10th place(07+57)=64
• 8 years agoHelpfull: Yes(0) No(2)