find the last two digits of the following expression

(1122)^(1122!)

• 76 is the answer
• 9 years agoHelpfull: Yes(11) No(2)
• 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
  
• 8 years agoHelpfull: Yes(9) No(0)
• 1122!=1122*1121*.....*4*3*2*1..so it can be written in 4*n form.so ques reduces to 1122^4...to find last two digits we have to devide it by 100..by remainder theorem:(1122*1122*1122*1122)/100..in each case remainder will be 22..so it becomes (22*22*22*22)/100-->(484*484)/100-->(84*84)/100-->7056/100-->56..so and is 56.
• 9 years agoHelpfull: Yes(6) No(9)
• Since power is 1122! so its remainder will be 2 when devided by 4.
  So question now reduces to find last two digits of 1122^2.(i.e 1258884)
  By remainder theorem,1258884/100--->84 will be remainder.so ans is 84
  
• 9 years agoHelpfull: Yes(1) No(8)
• ans is 22
  and correct ans...
  
• 9 years agoHelpfull: Yes(0) No(6)
• 4
  1122! has last digit 0 and 2^0 end digit is 2
• 9 years agoHelpfull: Yes(0) No(6)
• no of zeros in 1122!
  1122/5=224+44+8+1=277 zeros
  so 1122! is divisible by 4
  so unit place of 1122^1122!
  2^4=2
  so unit place must be 2
• 9 years agoHelpfull: Yes(0) No(2)