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Select 5 numbers. Using exactly five copies of each, simultaneously create 15 (multi-)sets, one of each sum from 1 to 15 inclusive.

Thanks to Gordon Hamilton for the inspiration...

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  • $\begingroup$ Despite a successful solution (though perhaps not the only one possible!), it seems that some confusion remains. Could you please let me know what needs clarification, so that I can improve the puzzle? Thanks... $\endgroup$ – Zomulgustar Dec 24 '18 at 6:28
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We must select 5 numbers with sum $\frac{1}{5}(1+2+\dots+15) = \frac{1}{5} \cdot 120 = 24.$

Select $1, 3, 5, 7, 8.$ Here is the list:

$1$

$1, 1$

$3$

$1, 3$

$5$

$3, 3$

$7$

$8$

$8, 1$

$5, 5$

$8, 3$

$7, 5$

$8, 5$

$7, 7$

$8, 7$

Motivation: $1$ is forced. I tried adding $2$ next, but that would've forced me to use a number above $8$ ($\min \max = 9$ achieved with $\{1, 2, 4, 8, 9\}$), so I added $3$ to the list. The only 5 element set with $1, 3$ and $\max \le 8$ was $\{1, 3, 5, 7, 8\},$ and it worked.

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  • $\begingroup$ Nicely done! The solution is unique if you restrict to natural numbers. $\endgroup$ – Zomulgustar Dec 23 '18 at 21:09

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