find the last two digit of 136^132!

• last two digit
  
  6^1=06
  6^2=36
  6^3=16
  6^4=96
  6^5=76
  6^6=56
  6^7=36
  6^8=16
  6^9=96
  6^10=76
  6^11=56
  
  repeatation of last two digit is after every 5 so 132!/5 is perfect divisible by 5 so there is no remainder so last two digit is 6^5=76
• 9 years agoHelpfull: Yes(2) No(2)
• 36
  the unit place is always going to be 6 for tens place just after going through several iterations the pattern is foun to be 3..53197..531975...so except the first no. rest is repeating with a cycle of 5 so iviing 131 by 5 remainer is 1 so tens place is 3 and unit place is 6
• 9 years agoHelpfull: Yes(0) No(4)
• 136^132!
  
  we can get the last 2 digits by dividing it by 100
  
  i.e------>136^132!%100--------->36^132!%100
  
  now by binomial theorem----36^132!=(6+30)^132!=6^132! + 132!*(6^(132!-1))*30^1+.......
  
  now if we see,from 2nd term itself every every term is divisible by 100
  
  so 6^132!%100------:
  
  6^1=06
  6^2=36
  6^3=16
  6^4=96
  6^5=76
  6^6=56
  6^7=36
  6^8=16
  6^9=96
  6^10=76
  6^11=56
  
  repeatation of last two digits is after every 5 cycles so 132!/5 is perfect divisible by 5 so last two digits are 6^5=76
• 7 years agoHelpfull: Yes(0) No(0)