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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 approach | Comment 3 of 4 |

Let the smallest number be n. Then, the numbers are n, n+2, n+5, n+6, n+8, n+10, n+11, with one of them being repeated. Let the repeat be n+x.

The smallest number obviously leaves out n+11. Hence, adding the other 7,

7n+31+x = 418 = 7*59+5.

=> 31+x = 5mod7

=> x = 2 0r 9 => x =2

=> 7n+33 = 418 => n = 55.

So, the numbers are 55, 57, 57, 60, 61, 63, 65, 66.

Nirmal


  Posted by nirmal on 2006-08-08 17:36:32
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