@sumit_agarwal I would urge you to consider a few examples yourself and try and work out first along the lines of above logic. Do it by standard approach first. It would help you develop a more generic intellect towards non-standardised questions like these. Feel free to comment again if it doesn't work out. Will try and provide a solution for your query then.

Let the marble numbers removed be (a + 1), (a + 2)… (a + n). Then, 1615 = 2n * (2n + 1)/2 – n * (2a + n + 1)/2 3230 = 2n * (2n + 1) – n * (2a + n + 1) 3230 = n * (4n + 2 – 2a – n – 1) = n * (3n – 2a + 1) 2 * 5 * 17 * 19 = n * (3n – 2a + 1) n = 34 or 38 satisfies. Sum = 72

Sum of all elements of set S = 465. For any subset having sum of elements x, there will be one complimentary set which is having sum of elements equal to (465 – x). Example let there be a set {1, 2, 3, 4}. Then subset which is having sum =3 will be {1, 2} {3}: (2) and subset having sum (1+2+3+4) -3 =7 will also be 2: {3,4} {1,2,4} So, number of subsets having sum of elements x = Number of subsets having sum of elements (465 - x) => No of subsets having sum 1 = No of subsets having sum 464 Also, No of subsets having sum 2 = No of subsets having sum 463 …. No of subsets having sum 232 = No of subsets having sum 233 Adding them we will get, No of subsets having sum less than or equal to 232 = No of subsets having sum more than 233 = 2^30/2 = 2^29

@anubhav_sehgal Hey! Can you please tell me that why is power of 3 inside the bracket 0? (In the question 58!-38!)

42 = 2 * 3 * 7 No of zeroes in 2006! will be the highest power of 7 in 2006! [2006/7] + [2006/49] + [2006/343] = 331 as 42 in base 42 will be 10 and will be the provider of trailing zeroes

In keynote 4, example 2, total non-negative unordered integral solutions is just 1 (4,0) and total non-negative ordered integral solutions are 2 (4,0 and 0,4). Please Correct me if I'm wrong.