find the last two digits in 1941^3483+1961^4181=

• rule : [any odd no's not ending with 5]^20m thn its last digit will be 01
  
  1941^3480 *1941^3+ 1961^4180*1961= 01*1941^3+01*1961
  
  1941^3=21 as last two digit and when add with 1961 we get 82 as last two digit so 82 is the last two digit..................
• 11 years agoHelpfull: Yes(14) No(6)
• the cyclicity of 1941 in units place is 1,1,1.... and 2nds place is 4,8,2,6,0 so last 2 digits are 21 and for 1961 units place is 1,1,1... and 2nds place is 6,2,8,4,0 hence last 2 digits are 61 add 21+61=82
• 11 years agoHelpfull: Yes(12) No(3)
• 82..
  do in calc :P
• 11 years agoHelpfull: Yes(1) No(4)
• sunnu explain it more properly
  
• 11 years agoHelpfull: Yes(1) No(0)
• 1941^3483+1961^4181
  For numbers in the form of (ab...c1)^(xy...z),the last digit is always one & 2nd last digit is last digit of c*z.
  eg:
  (c=4*z=3)=1(2.Consider 2 as 2nd last digit and last digit is 1
  21+61=82
• 10 years agoHelpfull: Yes(1) No(0)
• @sunnu can u give some more explanation???
• 11 years agoHelpfull: Yes(0) No(1)
• 3480/5 and but we need 3 more so 3rd place v have 2 similarly 4180/5 but 1 more so 1st place is 6
• 11 years agoHelpfull: Yes(0) No(2)