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A Distinct Sum Puzzle (Posted on 2006-07-28) Difficulty: 3 of 5
Determine a list of eight positive integers (not necessarily distinct) such that summing seven of them in all eight possible ways generates only seven distinct results: 418, 420, 423, 424, 426, 428 and 429.

See The Solution Submitted by K Sengupta    
Rating: 3.5000 (4 votes)

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another way | Comment 2 of 4 |

One sum , out of 8 possible  , is missing . Denote it by x

Let S8   be the sum of all  8  numbers and S7 of the 7 listed 

S7= 418+420+...+429=2968

Clearly  c+x= 7*S8     IS DIVISIBLE BY 7

SO X MUST BE DIVISIBLE BY 7, SINCE  2968 IS 

 X   HAS TO BE ONE OF THE NUMBERS LISTED  =>>  x=420

and S8=3388/7=484

NOW  THE 8 NUMERS ARE DERIVED:

484-418=66

484-420=64

484-420=64     THE EXTRA PARTIAL SUM

484-423=61

....ETC

484-429=55

,,,,,,NICE PUZZLE


  Posted by Ady TZIDON on 2006-07-29 01:43:52
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