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