last two digit 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
  so 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
  
• 11 years agoHelpfull: Yes(37) No(4)
• 1122! is an even no having a lot of zeroes at the end
  so lets take 1122 ^ 100 as example
  =22^100
  =(2*11)^100
  =2^100 * 11^100
  =(2^10)^10 * 01
  =76*01
  =76..
• 11 years agoHelpfull: Yes(17) No(2)
• 1122^1122! = 2^1122!*561^1122!
  = (2^10)even *(01) =76
• 11 years agoHelpfull: Yes(4) No(4)
• 2^(even num) always gives last two digits as 76
• 11 years agoHelpfull: Yes(2) No(9)
• we know that factor of any no greater than 1 is always even.
  So,22^1=22,22^2=484,22^3=10648,22^4=234256.Cyclicity of 22 is 2,4,8,6 for unit digit and 22,84,48,56 for last two digit.But we have to find 22^even no..So the last two digit is 56..
• 11 years agoHelpfull: Yes(0) No(7)
• there is one formula to count successive discount that is :- P(1+r1/100)(1+r2/100). Here r1 and r2 are % of discount. after outting values in this formula we can get all values that are (a)100.40 (b)156.95 (c)150 (d)145.35. so here least price is 100.40 so answer is (A).
• 11 years agoHelpfull: Yes(0) No(3)
• 76
• 11 years agoHelpfull: Yes(0) No(2)