A Distinct Sum Puzzle (Posted on 2006-07-28)

	Submitted by K Sengupta
Rating: 3.5000 (4 votes)
Solution:	(Hide)
The required whole numbers are 55, 56, 58, 60, 61, 64, 66 and 64. EXPLANATION: Subtracting the first result from the second result, third result, and so on – in turn, we respectively obtain 2, 5, 6, 8, 10 and 11 and accordingly seven out of the eight numbers to be determined must possess the respective forms P-2, P-5, P-6, P-8, P-10 and P-11, where P is a positive integer. The sum of these seven numbers is (7P - 42) which must be congruent to one of the seven given sums. Now, an integral solution for P is feasible only when, 7P -42 = 420; giving, P = 66. Consequently, seven out of the eight integers are equal to 55, 56, 58, 60, 61, 64 and 66. These seven integers add up to 420, which is not the minimum amongst the given sums. So, two of the eight numbers to be determined must possess the same value. Let this value be Q. Then, clearly we must have: Q+ 55+ 56+ 58+ 60+ 61+ 64 = 418, giving Q = 64. Consequently, the required numbers are: 55, 56, 58, 60, 61, 64, 66 and 64.

	Subject	Author	Date
	Acknowledgements	K Sengupta	2006-08-31 03:09:31
	Another approach	nirmal	2006-08-08 17:36:32
	another way	Ady TZIDON	2006-07-29 01:43:52
	Solution = spoiler	Leming	2006-07-28 16:01:00