An old man has Rs. (1! + 2! + 3! + ... + 50!), all of which he wants to divide equally (without fractions) among his n children. Then, n may be

• ans is 3
  there are three children for a old man
  1!+2!=3 so these Rs.3 can be divided 1 for each one
  on from 3!,4!,5!......50! every number is divided by 3
  for example 3!=6 ,Rs.2 for each
  for example 4!=24,Rs.8 for each
  and so on...
  so there three children for old man
• 10 years agoHelpfull: Yes(29) No(6)
• From 6! on wards 9 can divide dirctly.
  Becuse after 5! every number contains atleast two 3.
  Therefore
  The rem of 1! +2!+3!+4!+5! div by 9 = 0.
• 10 years agoHelpfull: Yes(7) No(3)
• in openSeesame options were -
  
  11,9,5,7
• 10 years agoHelpfull: Yes(1) No(0)
• 9;
  (1!+2!+3!+4!+5!+6!) is divisible by 9
• 10 years agoHelpfull: Yes(0) No(3)
• 1+2+3+3!(1+4+5*4+......)/3=0
  so n=3
• 10 years agoHelpfull: Yes(0) No(0)
• upto 5! sum is 120..divide it by 6...there after everyhing is divisible by 6 soo it can be shared without fractions among all the 6 sons equally..ans msy be 6 also
• 10 years agoHelpfull: Yes(0) No(0)