find last two digits of 1122^1122! ??

• here 2 in the unit place so
  2^10)^even=76
  so ans is 76
• 11 years agoHelpfull: Yes(15) No(4)
• (1122)^(1122!)=(2*561)^1122!=(2^10)^(1122!/10) * (561)^1122!
  last 2 digit of (2^10)^(1122!/10)= (XX24)^even no= 76
  last 2 digit (561)^1122!=01
  
  therefore last digit =76*01= 76
  ans 76
• 11 years agoHelpfull: Yes(10) No(4)
• last 2 digits of 1122! is 00.
  2^10=1024
  24^even=76
  so ans is 76
• 11 years agoHelpfull: Yes(3) No(2)
• (11*2)1122 then (2^10)112*2^2) which is 4 and 11^1122 is 21 so 21*4=84
  
• 11 years agoHelpfull: Yes(1) No(5)
• 76 ans is
  
• 11 years agoHelpfull: Yes(1) No(3)
• Can u explain dis plz
• 11 years agoHelpfull: Yes(1) No(1)
• 1122! : last two digits = 00.
• 11 years agoHelpfull: Yes(1) No(5)
• as it is 1122! in the power.. it will be 01
• 11 years agoHelpfull: Yes(1) No(2)
• 22^2=xx84.... power cycle of 84 will repeat after each 10 term... thus ans is 83.(bcz 1122! contains so many zeros i.e multiples of 10)
• 11 years agoHelpfull: Yes(1) No(0)
• ans is 76
  
• 11 years agoHelpfull: Yes(0) No(0)
• 76 is the answer.....
• 10 years agoHelpfull: Yes(0) No(0)