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You have found $13$ gold coins and strangely their weights are from $1$ to $13$ grams (such as $1,2,3,...$). You are bored and out of the blue you decided to divide golds into groups such as the sums of the weights of the golds in all groups will be the same.

In how many distinct ways can this be done?

If this question was asked for $7$ gold coins with $1$ to $7$ gram weights: The answer would be 5, such as (1-6-7, 2-3-4-5), (1-6,2-5,3-4,7), (1-2-4-7,3-5-6), (1-2-5-6,3-4-7), (1-3-4-6,2-5-7) etc.

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  • $\begingroup$ By "strangely their weights are changing" do you mean there are mysterious forces at work, so the coin weights are not always the same when weighed, or do you mean their weights are in the range 1 to 13 grams, or do you mean there is one coin of each weight? $\endgroup$ – Weather Vane Sep 6 '18 at 17:51
  • $\begingroup$ @WeatherVane check the example please $\endgroup$ – Oray Sep 6 '18 at 18:00
  • $\begingroup$ OK the weights are not changing, there is one coin of 1 gram, one coin of 2 grams etc? $\endgroup$ – Weather Vane Sep 6 '18 at 18:01
  • $\begingroup$ @WeatherVane yes exactly. edited accordingly. $\endgroup$ – Oray Sep 6 '18 at 18:01
  • $\begingroup$ This was an interesting question @Oray, especially given the numbers you have selected. $\endgroup$ – El-Guest Sep 6 '18 at 18:07
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The first thing to note is that

$1 + 2 + ... + 13 = 91$; and $91 = 13 \times 7$, so we must have either 13 groups of 7 or 7 groups of 13. Since we have coins heavier than 7 grams, the first option is out. We must therefore have 7 groups of 13.

The answer is therefore

1 way, since 13 must be alone; 12 must be with 1; 11 must be with 2; 10 must be with 3; 9 must be with 4; 8 must be with 5; and 7 must be with 6.

Then the groups are

{13, 12-1, 11-2, 10-3, 9-4, 8-5, 7-6}

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  • $\begingroup$ rot13(Ubj nobhg chggvat nyy bs gur pbhag va bar tebhc bs guvegrra pbvaf)? $\endgroup$ – EKons Sep 6 '18 at 18:14
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    $\begingroup$ @ΈρικΚωνσταντόπουλος thé question states “groups” which seems to imply plural or multiple groups (ie more than one). You’ve found the trivial solution! ;) $\endgroup$ – El-Guest Sep 6 '18 at 18:16
  • $\begingroup$ Yeah, not worth posting as a partial though, it's too trivial. $\endgroup$ – EKons Sep 6 '18 at 18:18

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