123^123! what is the late two digit

• 123^123!
  (123)^(----28 zeroes)
  consider only last two digit
  (23)^4)^(--000000000)
  (--41)^(---00000000)
  
  unit digit = 1 ,tens place = (tens of 41)*(unit of ---000)= 4*0=0
  
  => last two digits = 01
• 9 years agoHelpfull: Yes(25) No(4)
• 123! is obviously divisible by 4.
  By power rule of 3(123) the last number must be 1 (since last digit can be 3 is 3,9,7,1)
  therefore the last two digit should be X1.
  Now 123! have 28 zeroes(since |123/5|+|123/25| =28) therefore it must be 4*XXUUU..0000..00.. therefore the second last digit should be X*0=0.(by Eulers form)
  
  So the last two digits are 01
• 9 years agoHelpfull: Yes(8) No(0)
• ANSWER==>01
  as 123!= 123*122.......4*3*2*1= 4n number
  so last digit will be= 1 as 3^4 =81
  now 123! contains [123/5+123/25]=28 zeros
  so we can write 123!=..........000000
  2nd last digit will be = 2*0= 0(according to the Euler's rule 2nd last digit of power* 2nd last digit of the number= 2nd last digit)
  so last two digits = 01
• 9 years agoHelpfull: Yes(6) No(0)
• 3^1=1
  3^2=9
  3^3=27
  3^4=81
  after that all last digit is repeated by same number as 3 9 7 1
  so after 4! all values are get divisible by 4! so the reminder is 0.so the last two digit is 01!!
• 9 years agoHelpfull: Yes(0) No(2)