Let

*r*,

*o*,

*y*,

*g*,

*b*, and

*v* be the number of each marble colour.

By clues 1 and 6, *b* > 1, and by clues 5 and 6, *b* is a divisor of *r* and *b* < *r*. Therefore, by clue 4, *g*, *b*, and *v* are prime, and *r*, *o*, and *y* are not.

By clue 3, *r* < *o* and *r* < *y*. By clues 1 and 6, *g* < *b* and *v* < *b*.

Considering primeness and the orders established so far, *b* = 5, *r* = 6, *g* and *v* equal 2 and 3 in some order, and *o* and *y* equal 8 and 9 in some order.

By clue 2, *o* + *g* = 10. The only possibility is *g* = 2 and *o* = 8, and this means that *v* = 3 and *y* = 9.

The solution is *r* = 6, *o* = 8, *y* = 9, *g* = 2, *b* = 5, and *v* = 3.