# Dividing the first 10 primes into groups whose sum is prime [closed]

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 $$[2,10]$$ ?

• Wow I didn't think this is a textbook problem - I've never seen anything like it. Jan 26 at 7:50

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: -

• Correct and well done! Jan 25 at 3:55
• @DmitryKamenetsky I've updated the solution so that all groups are "in order", i.e. they consist of consecutive primes. Jan 25 at 4:02
• That is very sexy, thanks! Jan 25 at 4:24

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

• This is correct. Jan 25 at 3:56