What is the probability that a number 'N' will divide the sum of all numbers till 'N', for sufficiently large values of 'N'?

• sum of 1st n numbers = n(n+1)/2
  We have to check when n(n+1)/2 is divisible by n
  Case 1: When n is even
  n(n+1) will be divisible by n resulting in (n+1) and since n is even (n+1) being odd will not be divisible by 2.
  Case 2: When n is odd
  n(n+1) will be divisible by n resulting in (n+1) and since n is odd (n+1) being even will be divisible by 2.
  So when n is very large, we have almost equal no of even and odd numbers.
  So the probability is 1/2.
• 9 years agoHelpfull: Yes(24) No(0)
• Initially, even i thought so.
  But let us consider a case.
  n =100. sum = (100*101)/2 that is, sum = 5050.
  Which is not divisible by 100.
• 9 years agoHelpfull: Yes(4) No(0)
• Thanks kar,
  sum of n terms is n(n+1)/2 for this to be divisible by n, n+1 should be divisible by 2 and hence n has to odd . As for the number to be odd or even probability is 1/2, hence 1/2 is the answer.
• 9 years agoHelpfull: Yes(3) No(0)
• I applied the same reasoning Ashima.
  I am just checking it. I don't know the answer.
• 9 years agoHelpfull: Yes(1) No(0)
• take two consecutive even and odd nos and check
  lets take 20 & 21
  case 1:when n is even 20(20+1)/2=sum=>210/20=>10.5 not divisible
  case 2:when n is odd 21(21+1)/2=sum=>231/21=>11 its possible and hence probability 1/2;very simple if u go with values:)
• 9 years agoHelpfull: Yes(1) No(1)
• sum of n terms is n(n+1)/2 which will be always divisible by n so probability is 1.
• 9 years agoHelpfull: Yes(0) No(8)