the last two digits of (123)^(123!)

• 123! has lots of zeroes at the end so it is divisible by 4
  unit digit of (123)^(123!)=1
  10th digit of (123)^(123!)=unit digit of 123 * 10th digit of 123!
  =3*0=0
  hence last two digit of (123)^(123!)=01
  
• 10 years agoHelpfull: Yes(14) No(0)
• ans:01..
  first u know 123! have so many zeros...123/5...atlst 24...now we can easily take out 2 or 4 from this mnimum 24 zeros..
  now take 123^2.....
  as we know anything end with 3 we hv to tk it sqr rt..
  ok 123^2 ends with 23^2..as 29..
  no take 9 s power cycle and we got 29^4..as 29^2*29^2...ends with 50-29=21^2 as 41*41..ends with 50-41=9^2..as 81..
  now u know a number end with 1^1245566.....ends with 1...and last two digit is very easy to count...so 81^mnoop0000000000 gives us 01 at end..
  
  
  now the trick if there is a factorial in the power which is huge..u frgot about that..
  then take out 4and 2 frm it..
  and u can do it,,
  so again frget it..
  ur moto is to end the base one end with 1..which ends in 01 as a last two digit..
  
• 10 years agoHelpfull: Yes(3) No(3)
• (123!)= ...............................00 its last two digits are 0s
  and our question becomes 123^.......00
  =__3^......00
  =__9^(....00/2)
  =__(-1+10)^(.....00/2)
  by apllying binomial fomula we will get the 1 as first term so the last two digit will be 01
  ans= 01
• 10 years agoHelpfull: Yes(2) No(0)
• 01 ans.
  
  
  
  
  
• 10 years agoHelpfull: Yes(1) No(0)
• 123^123!= (3*41)^123!=(3^4)(123!/4) * (41)^123!= last 2 digit=01
  3^4=81 (81)^(123!/4)=01
  41^123!=01
  
  thus 01*01=01 ans 01
• 10 years agoHelpfull: Yes(1) No(1)
• ans is 01
  
• 10 years agoHelpfull: Yes(0) No(1)