Latest Elitmus Aptitude Question SOLUTION: What is the probability that a number 'N' will divide the sum of all numbers till 'N', for sufficiently large values of 'N'?

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(#M40151820) ELITMUS Numerical Ability question Probablity Keep an EYE

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

Asked In Elitmus KAR (9 years ago)

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Pallavi Gupta  (9 years ago)

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.

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KAR  (9 years ago)

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.

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Ashima Sabharwal  (9 years ago)

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.

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KAR  (9 years ago)

I applied the same reasoning Ashima.
I am just checking it. I don't know the answer.

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manikandan  (8 years ago)

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:)

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Ashima Sabharwal  (9 years ago)

sum of n terms is n(n+1)/2 which will be always divisible by n so probability is 1.

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