What are the last two digits of expression 5306^214?

• 96
  hence, 6^214=6^200*6^14
  =(6^2)^100*(6^2)^6*6^2
  =>>6*6*36
  as the expansion of (6^2) must consist 36
  so last 2 digit =>>96(answer)
• 8 years agoHelpfull: Yes(10) No(0)
• 24
  multiply last 2 no.s 6*4=24
• 8 years agoHelpfull: Yes(3) No(9)
• 5306=2*7*379
  So (5306)^214=(2*7*379)^214
  2^214=(2^10)^21 *2^4 = (1024)^21 * 16 = 24*16=384
  7^214=(7^4)^53 * 7^2 =(49*49)^53 * 7^2
  =(2401)^53 * 49 = 01*49=49
  (379)^214= (79)^214 = 21^214 =81
  
  84*49*81=96
• 8 years agoHelpfull: Yes(2) No(0)
• 06 would be the answer
• 8 years agoHelpfull: Yes(1) No(6)
• take the lost two digits and raise to power
  41^3=68921
  61^1=61
  add both lost digit will be 82.
• 8 years agoHelpfull: Yes(0) No(1)
• 96 would be the ans
• 8 years agoHelpfull: Yes(0) No(0)
• ans will be: 06
  0*4= 0
  6^4 = 6
• 8 years agoHelpfull: Yes(0) No(1)
• CORRECT ANSWER IS 96
  CYCLICK ORDER OF 6 IS 5
  SO 214/5, REM = 4
  ie 6^4 = 96
• 8 years agoHelpfull: Yes(0) No(2)
• 76 power any number is 76 and multiplication of 76 with any 6 multiples ending with 6 gives same last two digits.
  eg:
  76^1 = 76
  76^2 = 5776
  76^3 = 438976 ....... last two digits are 76 only
  and
  76*36 = 2736 ,76*96 = 7296....
  6^214 = 6^(5*42)*6^4 = 76*96 = 96 (since 6^5 gives 76)
  last two digits are 96.
• 6 years agoHelpfull: Yes(0) No(0)