Glycereine

I'll give someone else a chance at the winners . I have a theory though...

Thanks!! I encourage others to do it though without looking, it doesn't take too long but you can make it harder than it should be (I almost did)

[spoiler='Proof ']First, Q,S,U,W and Y must be even numbers (they are the second digit of numbers that must be divisible by a number with a factor of 2. Therefore the others must be odd. - Y must = 0 as any number divisible by 10. - T must = 5 as any number divisible by 5 must end in 0 or 5 (0 is taken by Y) - U must = 4, for working down, the only number divisible by 6 that starts with 5 is 54 - V must = 9 for the only two numbers divisible by 7 that begin with 4 are 42 and 49, but V must be odd (see top) - W must = 6 as the only number beginning with 9 that is divisible by 8 is 96 - X must = 3 because the only number divisible by 9 beginning with 6 is 63 Simple part is done. Now starting back at S, the lowest digit we do not yet know. RS must be divisible by 4 and the only 2 even numbers remaining are 8 and 2. S cannot be 8 because the two odd numbers left, 1 and 7 combined with 8 are not divisible by 4. 18/4 = 4.5 and 78/4 = 19.5 - S must = 2 - Q must = 8 as it is the only even number left - R can be either 1 or 7 because both 12 and 72 are divisible by 4. Also both 81 and 87 are divisible by 3. - P can be anything because no matter what the number starts with PQ will be divisible by 2 if Q is even. Therefore P = 7 or 1 (as well as R = 1 or 7). Swapped with each other they make the two possibilities

Unfortunately I just don't get what the patches would look like so I guess I'm out on this one.

Additionally you said that likely A and E will die in the first round which is not possible. If E dies in the first round, no one can shoot at A because D has already fired his shot (killing E).