I'm trying to word the following mechanic, so that it fits on half of a poker sized card. Fewer words the better :)

There are 2 dice, call maneuver dice, that are rolled. There is a card in your hand that has a die (1 - 6) on it. You can play the card if you can figure out a way to add and subtract, in any order, the two maneuver dice and the die on the card so that the total is divisible by 3. You have to use all three dice.

for example: you maneuver dice are 5 and 2. the die on the card is a 1. To play the card, you could use the following math:

5 - 1 + 2 = 6. 6 is divisible by 3.

Here's my current attempt:

Play this card when adding and subtracting, in any order, the two maneuve dice and this card results in the total being divisible by three.

Thanks for the input.

Some background:

This is for my gladiator game. I've been trying to come up with a way to entertain the players not currently fighting (fights are only between two players). I thought I would try this little mini game. The two players fighting have a maneuver die. The players watching would draw 3 observation cards (with the die on it), and watch the fight. During the fight the maneuver dice are maniplated by different moves the gladiators can perform. The maneuver dice usually increase in value (six being the max). The players watching would play an observation card (one per round. There are multiple rounds per fight) when the condition on the card were met. The goal would be to make 5 observation before the fight was over. If the goal was met they would get a bonus for their gladiator's next fight.

Thoughs on comments:

Now that I know that the cards can be "played unless exactly two dice are 3 and 6", the solution to the mini game doesn't take much thought and is not as interesting to me. O'well, back to the drawing board. I'm glad we have some mathematicians here, I don't think I would have figure that out on my own.