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Take the first 10 primes. Can you divide them into $g$ disjoint groups, such that the sum of numbers in each group is prime. In particular can you make this work for every value of $g$ in the range $[2,10]$ ?

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  • $\begingroup$ Wow I didn't think this is a textbook problem - I've never seen anything like it. $\endgroup$
    – Dmitry Kamenetsky
    Jan 26, 2021 at 7:50

2 Answers 2

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Below is one solution. Groups of 1 are implied and not shown


2: sum 3...29 = 127
3: 2+3 = 5, 5+7+11 = 23, 13+17+19+23+29 = 101
4: 5+7+11 = 23, 13+17+19+23+29 = 101
5: 2+3 = 5, 5+7+11 = 23, 19+23+29 = 71
6: 5+7+11 = 23, 19+23+29 = 71
7: 2+3 = 5, 5+7+11 = 23
8: 5+7+11 = 23
9: 2+3 = 5
10: -

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  • $\begingroup$ Correct and well done! $\endgroup$
    – Dmitry Kamenetsky
    Jan 25, 2021 at 3:55
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    $\begingroup$ @DmitryKamenetsky I've updated the solution so that all groups are "in order", i.e. they consist of consecutive primes. $\endgroup$
    – Paul Panzer
    Jan 25, 2021 at 4:02
  • $\begingroup$ That is very sexy, thanks! $\endgroup$
    – Dmitry Kamenetsky
    Jan 25, 2021 at 4:24
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If I understood the problem correctly. One of the answers is:

G10. 2,3,5,7,11,13,17,19,23,29
G9. 3,5,7,11,13,17,19,23,29+2
G8. 2,3,5,7+11+13,17,19,23,29
G7. 3,5,7+11+13,17,19,23,29+2
G6 2,3+5+11,7+13+23,17,19,29
G5 3+5+11,7+13+23,2+17,19,29
G4. 2,11,7+13+23,17+19+29+3+5
G3. 3+5+11,7+13+23,2+17+19+29
G2. 2,3+5+7+11+13+17+19+23+29

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  • $\begingroup$ This is correct. $\endgroup$
    – Dmitry Kamenetsky
    Jan 25, 2021 at 3:56

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