find last two digits of (1941^3843)+(1961^4181)

• 1941^3843
  first take last two digit 41
  
  first last digit is 1 ..............(1)
  then take the 4 (1941) and multiply with 3 (3843)
  4*3=12,take last digit from this is 2 ...........(2)
  now (1)(2)
  so last two digit is 21
  
  now same for 1961^4181
  first take last two digit 61
  
  first last digit is 1 ..............(1)
  then take the 6 (1961) and multiply with 1 (4181)
  6*1=6,take last digit from this is 2 ...........(2)
  now (1)(2)
  so last two digit is 61
  last two digits of (1941^3843)+(1961^4181) is 61+21=82 (ans)
• 12 years agoHelpfull: Yes(33) No(2)
• 82..
  last 2 digit of (1941^3843)is last 2 digit of (((1941^3840)*41)*41)*41)i.e, 21. in this way last 2 digit of (1961^4181) is 61. so 61+21=82.
• 12 years agoHelpfull: Yes(14) No(5)
• When an integer with an ODD UNITS DIGIT OTHER THAN 5 is raised to A POWER THAT IS A MULTIPLE OF 20, the last two digits of the result are 01.
  
  Thus, the last two digits of 1941³⁸⁴⁰ and 1961⁴¹⁸⁰ are 01.
  To determine the last two digits of 1941³⁸⁴³ and 1961⁴¹⁸¹, keep multiplying the last two digits by 41.
  
  Last two digits of 1941³⁸⁴³:
  1941³⁸⁴⁰ --> 01.
  1941³⁸⁴¹ --> 01*41 = 41.
  1941³⁸⁴² --> 41*41 = 1681.
  1941³⁸⁴³ --> 81*41 = 3321.
  Thus, the last two digits of 1941³⁸⁴³ are 21.
  
  Last two digits of 1961⁴¹⁸¹:
  1961⁴¹⁸⁰ --> 01.
  1961⁴¹⁸¹ --> 01*61 = 61.
  Thus, the last two digits of 1961⁴¹⁸¹ are 61.
  
  Thus:
  (last two digits of 1941³⁸⁴³) + (last two digits of 1961⁴¹⁸¹) = 21+61 = 82.
• 10 years agoHelpfull: Yes(7) No(1)
• 82 is d correct ans.
• 12 years agoHelpfull: Yes(5) No(1)
• plz elaborate....
• 12 years agoHelpfull: Yes(5) No(1)
• anirudh explain
  
• 12 years agoHelpfull: Yes(2) No(2)
• ans is 42?
• 12 years agoHelpfull: Yes(1) No(12)