Let

*r* be the number of red marbles,

*o* orange, etc.

The lowest number of marbles avoiding adjacent primes would be 2 + 5 + 11 + 17 + 23 + 31 = 89. This is close to the maximum so there will not be that many possibilities.

If we go to the next higher possibility, that is 2 + 5 + 11+ 17 + 23 + 37 = 95.

Can we increase the second-highest number? No, as the resulting sum would be 101.

We can increase the highest number again which gives us 2 + 5 + 11+ 17 + 23 + 41 = 99.

These are the only possibilities before we consider the other clues.

Since the total number of three marbles is one different from the total number of the other three marbles, we are looking for very particular subsets.

For 2 + 5 + 11+ 17 + 23 + 41 = 99, the subsets sum to 49 and 50. 41 must be in one of the two subsets. The other two numbers must then sum to 8 or 9. With only 2 and 5 to choose from (the others being too large), obviously, there is no solution here.

For 2 + 5 + 11+ 17 + 23 + 37 = 95, the subsets sum to 47 and 48. 37 must be in one of the two subsets. The other two numbers must then sum to 10 or 11. With only 2, 5, and 11 to choose from (the others being too large), obviously, there is no solution here.

For 2 + 5 + 11+ 17 + 23 + 31 = 89, the subsets sum to 44 and 45. 31 must be in one of two subsets. The other two numbers must then sum to 13 or 14. With only 2, 5, and 11 to choose from (the others being too large), obviously, the subsets are {2, 11, 31} (sum 44) for *r*, *y*, *v* and {5, 17, 23} (sum 45) for *o*, *g*, *b*.

Since *y* > *g* and *v* > *g*, neither *y* nor *v* can be 2. They must be 11 and 31 in some order. Therefore, g = *5* and *r* = 2.

Since *b* < *o* and *b* < *v*, it must be that *b* = 17 and *o* = 23 and *v* = 31. This leaves *y* = 11.

There are 2 red, 23 orange, 11 yellow, 5 green, 17 blue, and 31 violet marbles.