(1141)^3843+(1961)^4181

• 41^3=21 last two digits and 61^1 last two digits =61 so 61+21=82
  so 82 is crrct
• 9 years agoHelpfull: Yes(23) No(3)
• unit digits will be 1 fr both
  n tens digit will be 4*3=2 and 6*1=6 resp.
  so 21+61=82
• 9 years agoHelpfull: Yes(17) No(0)
• unit place =1+1 = 2
• 9 years agoHelpfull: Yes(4) No(5)
• Question do not have sufficient data.Actual question is to find last two digit of this no.
  1141 last digit is 1
  4*3(where 4 is second last of 1141 and 3 is last of 3843)=12 we take two
  so no. is 21
  same 1961 have 1
  and 6*1=6
  so no. is 61
  total 21+61=82
• 9 years agoHelpfull: Yes(4) No(1)
• the unit place of (1141)^3843 is 1
  and the unit place of(1961)^4181 is 1
  so, the unit place of (1141)^3843+(1961)^4181 will be (1+1)=2
  
• 9 years agoHelpfull: Yes(3) No(5)
• they asked last 2 digits or the entire value??
  not possible to find entire value by human mind... :D
• 9 years agoHelpfull: Yes(1) No(5)
• unit digits will be 1 fr both
  n tens digit will be 4*3=2 and 6*1=6 resp.
  so 21+61=82
• 9 years agoHelpfull: Yes(1) No(0)
• ((4*3)*10+1) + ((6*1)*10+1)
  121 + 61= 182
  so the last 2 digit will be 82
  search for necessary option.
• 9 years agoHelpfull: Yes(0) No(2)
• unit digit power any thing will have last digit as 1 hence unit digit will be 1
• 9 years agoHelpfull: Yes(0) No(0)
• last 2 digit 82
• 9 years agoHelpfull: Yes(0) No(0)
• 4 16 1-----4641611124 10 1the last digit will be 01
• 9 years agoHelpfull: Yes(0) No(1)
• 41*3=21
  61*1=61
  61+21=81
• 9 years agoHelpfull: Yes(0) No(1)