what is the probability of divisibility of sum of first n number by n?

• 1/2
  every alternate sum of first n natural numbers is divisible by the number
  sum of first 'n'number=n(n+1)/2
  For n=2,3,4,5,6,7,8,...., sum=3,6,10,15,21,28,36,...respectively
  so probability=one among the two= 1/2
• 11 years agoHelpfull: Yes(28) No(0)
• we know that the sum of first n number is n(n+1)/2.if n is even than the probability will be 0. and if n is odd than the probability will be 1. if we r taking for first n than it will be 1/2.
• 11 years agoHelpfull: Yes(3) No(0)
• 0.5
  
  bcoz for odd its 1 and for even its 0 so overall probab = 1
  
• 11 years agoHelpfull: Yes(0) No(0)
• IT WAS GIVEN THAT N IS SUFFICIENTLY LARGE
• 11 years agoHelpfull: Yes(0) No(0)
• IF N IS LARGE i.e N(N+1)/2N WHICH IS (N+1)/2 WHICH CANT BE DETERMINED
• 11 years agoHelpfull: Yes(0) No(1)