what are the last two digits of expression 5306^214

• for 5306^214
  the last digit will always determined by the expansion of 6^214
  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)
  
  
  
• 9 years agoHelpfull: Yes(8) No(1)
• last two digit is 96...
  because cyclic factor of 6=1 that's why last digit =1
  and (5306*5306) last two digit is 36 and the power become 107...
  after 5 time multiplication we will get again 36..
  so 107/5 reminder=2
  36*6=216{we multiply by6 because in5306 second last digit is 0.it has no effect}
  16*6=96...this is the asn...
  
  
  
• 9 years agoHelpfull: Yes(4) No(0)
• easy method as unit digit is 6,,,214/6 we get rem as 4 so (6^4)=1296 the last two digit is 96.
• 9 years agoHelpfull: Yes(4) No(0)
• Answer is 06.
  
  Cyclic factor of 6 = 1 thats y last digitt =1.
  
  to fond the 10th's digit we have to multiply the 10th's digit from 5306 and the unit's digit from 214 which is 0. therfore 06 is the answer.
• 9 years agoHelpfull: Yes(1) No(0)
• 5306*5306=last two digit 36.after multiplication of five times it again 36.then 5036^210 last two digit 36.after rest 4 times it is 56.then 56 is answer
• 9 years agoHelpfull: Yes(0) No(6)
• when ever we have something as 6^x we will always get 6 as the last digit
• 9 years agoHelpfull: Yes(0) No(2)