What will be the last two digit of 1122^1122!

• 1122! have more than 2 two's hence it is a multiple of 4..
  now just calculate the last two digits of 1122^4
  it will be 56( by remainder theorem)
  since 1122! has ...000 at the end . Therefore it is a multiple of 10
  So,
  56^1= ...56
  56^2= ...36
  similarly,
  56^5=....76(last two digit only)
  now, 56^10=...76(last two digit only)
  Hence last two digit of 1122^1122! will be 76.
• 10 years agoHelpfull: Yes(20) No(20)
• (2^10)^even=76
  1121! is a even no
• 10 years agoHelpfull: Yes(11) No(9)
• answer is 84
  1122^1122== (102x11)^1122
  --> (((102^10)^112)x2^2 x 11^1122)
  -->(24^even x 2^2 x 21)
  ---> 76x04x21-----> 84.......awesome
• 10 years agoHelpfull: Yes(10) No(21)
• 1122^40*28+2....so last 2 digit=(1122)^2= ans is 84
• 10 years agoHelpfull: Yes(9) No(21)
• 84 cannot be the answer.
  because if we calculate the unit digit of the same problem then it is coming 6
  84 is the wrong answer..
• 10 years agoHelpfull: Yes(9) No(9)
• (1122)^1122!=(22)^1122!=(11)^1122! * (2)^1122!
  11^1122!=01
  2^1122!=76
  So,76*01=76
• 10 years agoHelpfull: Yes(6) No(10)
• If there is a base ending with 2, then each 20th power will end with 76 as last two digits. so upto 1122^1120, last two digits would be 76. and for 1122^1122,
  ........76 * 1122 gives .........72
  ........72 * 1122 gives .........84.
  So answer is 84.
• 9 years agoHelpfull: Yes(4) No(4)
• infinity ! :D
• 10 years agoHelpfull: Yes(3) No(15)
• 1122! have more than 2 two's hence it is a multiple of 4..
  now just calculate the last two digits of 1122^4
  it will be 56( by remainder theorem)
  since 1122! has ...000 at the end . Therefore it is a multiple of 10
  So,
  56^1= ...56
  56^2= ...36
  similarly,
  56^5=....76(last two digit only)
  now, 56^10=...76(last two digit only)
  Hence last two digit of 1122^1122! will be 76 :)
  
• 10 years agoHelpfull: Yes(3) No(4)
• 4 multiply it
• 10 years agoHelpfull: Yes(2) No(4)
• 1122^1122
  ==1122 can be written as 102*11
  ==(102*11)^1122
  ==(102^1122)*(11^1122) /*11 power anynuber results 21 as last two digits*/
  ==(102^(1120+2))*(21)
  ==(102^10^112)*(102^2)*(21)
  ==(24^112)(04)(21) /* here 102^2 gives 04 as last two digits and 102^10 gives 24 as last two digits 2^10 is 1024*/
  ==76*04*21 /* here 24 ^even is always 76*/
  ==....84
  
• 9 years agoHelpfull: Yes(1) No(5)
• 44
  2*2 lst digit
  2nd last 2*2
• 9 years agoHelpfull: Yes(1) No(1)
• 4 is the answer
• 10 years agoHelpfull: Yes(0) No(7)
• according to arun sharma book if 1122! is at power then divide it with a and we get remainder of 2 and last digit is 2 in unit place of base so unit digit is of 4
• 9 years agoHelpfull: Yes(0) No(3)
• unit place 2^2=4;tens digit 2*2=4
  so ans. is 44
• 9 years agoHelpfull: Yes(0) No(2)
• Answer will be "04"
  
  See it has a concept as: when there is base ending with 2, every 20th power will give 76 as last two digits
  from 21st power same cycle repeats upto 40th power
  hence till 1120, last two digits will be 76, then 1121->"02"
  then 1122->"04"
• 9 years agoHelpfull: Yes(0) No(3)