what is the last two digits of 1122^1122!

• 1122^1122!
  =(2*561)^1122!
  =2^1122!*561^1122!
  =(2^10)^1122!/10 * 561^1122!
  =(xx24)^1122!/10 * 561^1122!
  we know that, 24^even = xx76
  Last two digit of (xx24)^1122!/10 =76 (as 1122!/10 is even)
  Now last two digit of 561^1122!=01(unit digit is always 1,10th digit=10th Digit of 561*unit digit of 1122!=6*0=0)
  Hence last two digit of 1122^1122!=76*01=76
• 10 years agoHelpfull: Yes(15) No(4)
• 76 it is
  2^10^even will end with 76
  Now 1122! Will be an even number only...taking 10 common will result in even number...
  Thus 2^10^even= 76
• 10 years agoHelpfull: Yes(8) No(1)
• cyclicity of 2 is 4
  factorial value will be divisible by 4 and remainder is 0.
  number at the place of 0 in the cyclicity is 6
• 10 years agoHelpfull: Yes(5) No(3)
• 
  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(5) No(0)
• 1122*(1000+122)
  =1122000+1122*122
  1258884
• 10 years agoHelpfull: Yes(4) No(11)
• ans will be 76
• 10 years agoHelpfull: Yes(3) No(1)
• 76 coz 2^10^even and 1122! will be even.
• 10 years agoHelpfull: Yes(1) No(1)
• 84 last two digits means 22^1122
  22 we can write as (2*11)^1122 2^1122*11^1122= 4*21=84
• 10 years agoHelpfull: Yes(1) No(2)
• any even no raised to the power of 20k ends with 76 as its last two digit and any odd no raised to the power of 20k ends with 01 as its last two digit.. since 1122! will be a multiple of 20 (1*2*......*19*20*.......*1122) and 1122 is an even no therefore the ans will be 76.
• 10 years agoHelpfull: Yes(1) No(1)
• May be 6 I guess..
  
• 10 years agoHelpfull: Yes(0) No(5)
• 1122!is divibile by 4n so
  2^4 = 6 last digit is 6
• 10 years agoHelpfull: Yes(0) No(2)
• 16
  1122=11*102
  11^1122! * 102^1122!
  1122! contains some zeroes at the end..so 11^1122!=01 (last 2 digits)
  1122! is divisible by 4 so we are now left with (102)^4=16 (last 2 dgts)
  
• 10 years agoHelpfull: Yes(0) No(1)