find the smallest number such that n! is divisible 990

• The factor of 990 is : 11*5*3*3*2
  That why the maximum factor of 990 is 11 .
  so, n=11
  That means n!=11!= (11*10*9*8*7*6*5*4*3*2*1)
  And then n!/990 = (11!/990)= (11*10*9*8*7*6*5*4*3*2*1)/(11*5*3*3*2)
  So Ans is 11 .
• 9 years agoHelpfull: Yes(12) No(3)
• The factor of 990 is = 11*5*3*3*2
  =11*5*3^2*2
  maximum factor of 990 is 11 .
  to divisible by 990 n must be multiple of 11..
  here we required smallest no that's why n=11..
  so, n=11
  And then n!/990 = (11!/990)= (11*10*9*8*7*6*5*4*3*2*1)/(11*5*3*3*2)
  So Ans is 11 .
• 9 years agoHelpfull: Yes(1) No(1)
• 990=11*9*10
  Smallest number is 11! so that it can be divisible by 11,9&10.
• 9 years agoHelpfull: Yes(0) No(0)
• ans is 11
  because 990=2 * 3^2 * 5* 11
  so number should contain at least 11
  and 11! contains other factors also....
• 9 years agoHelpfull: Yes(0) No(0)