If mn=1024 and k=(1025!)/m!^n and k has 128 and more zeroes.what is the minimum value of n such dat k is +ve natural no.

Dnt knw the exact questn.

• m=1024
  n=1
  then k =1025
• 11 years agoHelpfull: Yes(14) No(8)
• k=(1025!)/m!^n where m*n=2^10
  
  1025! has 255 zeros, k can have 128 zeros so m!^n can have a maximum of 128 zeros.
  
  Now let m=2^3 and n=2^7
  m! has 1 zero
  m!^n will have 2^7 zeros ie 128 zeros that is not possible..
  
  Let m=2^2 and n=2^8
  m! has no zeros so m!^n will have no zeros..
  so least possible value for n is 2^8.
• 11 years agoHelpfull: Yes(13) No(1)
• can u explain d process akshay kanthed.....
  
• 11 years agoHelpfull: Yes(0) No(0)
• visit this for solution http://www.onlineadhyayana.com/ elitmus section...:)
• 11 years agoHelpfull: Yes(0) No(11)
• can u explain? rishabh singh bisht
• 9 years agoHelpfull: Yes(0) No(0)