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If the array list consists of all the numbers until n. Sum of all numbers including n would be n(n+1)/2. Now, we want to find 'a' subset whose sum is a multiple of n.
- varsrm November 13, 2014When n is odd, n+1 is even, hence, the sum of all the numbers including n would be a multiple of n. Hence, the answer here is a subset with all the numbers including n.
When n is even, sum of all even numbers until and including n is (n)(n+2)/2 which again is a multiple of n. Hence, the answer here is a subset with all the even numbers until and including n.
Same approach can be followed for multiples of 2N.