Let n be a natural number, then n(n*n-1) is always divisible by

• n(n*n-1)
  = n*(n^2-1)
  = n*(n-1)*(n+1)
  = (n-1)*(n)*(n+1)
  product of 3 conseutive natural no. is always divisible by 6.
  
• 10 years agoHelpfull: Yes(43) No(2)
• rakesh bro have u done PHD in numbers .:P
• 10 years agoHelpfull: Yes(22) No(0)
• by hit and trial method,
  firstly put value of n=2 then
  2(4-1)=6
  now put value of n=3 then
  3(9-1)=24
  n=4 then
  4(16-1)=60
  therefore in each case the result is divisible by 6
  answer is 6.
• 10 years agoHelpfull: Yes(6) No(2)
• 2
  hav som random examples trail and error
• 10 years agoHelpfull: Yes(1) No(0)
• 2 , 3 , 6 are the answers
  
• 10 years agoHelpfull: Yes(1) No(0)
• n-1, n, n+1
  any of the three consecutive natural number.
• 10 years agoHelpfull: Yes(0) No(0)
• n*(n-1)*(n+1) which is divisible by 6
• 10 years agoHelpfull: Yes(0) No(0)
• SINCE A NO. WHOSE MULTIPLE ARE n*(n-1)*(n+1)
  if n is odd or even then it must contain two multiples of two
  so its ans will be 4
• 10 years agoHelpfull: Yes(0) No(0)