plz correct answer he btana ...

find the last two digits of the following expression

(1122)^(1122!)

• (1122)^(1122!)=(11)^(1122!)*(102)^(1122!)
  1122!= will have a even number and zeroes for example:2536200000000
  (11)^(1122!)----will have last two digits as 01
  because it repeats itself every power of 10 like 11^10,11^20
  and so on......last two digits for all these are 01
  (102)^(1122!)---will have last two digits as 76
  bez it repeats every power of 20 like 102^20,102^40
  and so on ...has last two digits for all these are 76
  so 01*76=76
  so Last two digits of the expression is 76
• 10 years agoHelpfull: Yes(16) No(0)
• exlain to krdo
• 10 years agoHelpfull: Yes(6) No(1)
• 1122!=*****00
  So last digit is 2^4=*6(As remainder of the value divided by 4 is 0)
  And Last 2nd Digit is 2*0(As last digit is 0)
  last 2 digit is 06
  
• 10 years agoHelpfull: Yes(4) No(4)
• answer is 76
• 10 years agoHelpfull: Yes(2) No(0)
• 1122! is 10*even no from
  so (2^10) ^even always give last digit as 76
• 10 years agoHelpfull: Yes(2) No(2)
• Ans. iz 56
• 9 years agoHelpfull: Yes(2) No(1)
• when result of 28 sep kolkata test ll be declared, any idea guys??????
• 10 years agoHelpfull: Yes(1) No(3)
• answer is 76
  
• 10 years agoHelpfull: Yes(0) No(1)
• 1122!= will have a even number and zeroes for example:2536200000000...explain it @Ram or anyone else
• 8 years agoHelpfull: Yes(0) No(0)
• (1122)^(1122!)
  we can split 1122! in term of 4x+c
  there for
  1122!=4x+2
  1122^2= 125884
  
  so the last 2 digit of the number =84
• 8 years agoHelpfull: Yes(0) No(0)
• correct ans is 84 . go for it :D
• 8 years agoHelpfull: Yes(0) No(0)
• 1122/100 = 22 ^1122/100
  
  Now R1= 2^1122! /100
  
  R2= 11^1122! / 100
  
  1122! have many zeros in last … as w.k.t after 5! There may have one single zero
  
  So( 2^10)^even no / 100 = 76
  e.g 2^10=1024
  (1024)^2 mod 100 = 76
  (1024)^4 mod 100 = (24)^4 mod 100= 76
  (1024)^6 mod 100 =(24)^6 mod 100 =76
  so (2^10)^even always gives 76
  
  now 11^1122! /100
  see 11^2 = 121
  11^3 = __ 31 ( last two digit)
  11^4 = … 41
  Similarly 11^10 = 01 (last two digit )
  Now (11^10) ^even num / 100
  = 01 as last two digit
  So last two digit = R1*R2 = 76*1 = 76
• 7 years agoHelpfull: Yes(0) No(0)
• We know the last two digits can only result in a number lower than 100,
  we know multiplication between two large numbers is basically an accumulation of simple multiplications.
  In this case we have (1000+100+20+2)*(1000+100+20+2),
  since we can only allow numbers smaller than 100 we can drop 1000 and 100.
  We get (20+2)*(20+2) which results in 400+40+40+4,
  400 is greater than 100,
  so 40+40+4=84.
  84 are the last two digits.
• 7 years agoHelpfull: Yes(0) No(0)
• 1122*1121*1120!=last digit is 0
  and when it is expanded to 1100!,ssecond last digit is also 0.
  so answer is 00.
• 7 years agoHelpfull: Yes(0) No(1)