TheDoberman wrote:
Not a problem.
You are right in that drawing 10 Even cards gives you an even sum, as does 10 Odds. But what about if he draws the following 10 cards:
1, 2, 3, 4, 5, 6, 7, 8, 9, 10: The total is 55 (Odd sum = bad).
Another group of 10 cards:
1, 2, 3, 4, 5, 6, 7, 8, 9, 11: The total is 56 (Even sum = good).
Looking at this as a permutation problem will just drive you crazy - far too many options and I'm not sure it can get you the answer.
The way I read the question, the key word is "ensure" and I assume that he stops when he reaches an even sum - the wording isn't great. Based on the logic of my first post, there is no way you can draw 12 cards and not have achieved an Even sum at somepoint along the way (the worst case being that it happens on the 12th card). Note that you can draw 12 cards and get an odd sum (like 1,3,5,2,4,6,8,10,12,14,16,18 = 99) but along the way he'd gotten his even sum and stopped (1,3 = 4).
Did that help?
I'm not sure if your responding to my post, but I figured I would explain my reasoning. Again in this problem we want to know when he will know for sure when he will have an even number. We have to figure out the maximum amount of cards he could draw before he would have to draw an even amount. As my post described above, it would require that he the first card he draws is odd then, if he draws all the even cards subsequently the sum will always be odd. After you have exhausted your supply of 10 even cards the only thing you have left to draw would be an odd. Any odd plus and odd would be equal to even. 1+10+1=12, thus, this tells us that he can safely conclude that at 12 cards he will have drawn an even sum.
You really only need the properties of odd+odd, even+even, and odd+even to figure out this problem. You only need logic. If you are doing any permutations or combinations you are doing too much work.
You are correct in saying that it is impossible to draw 12 and not get an even, but 12 is the point at which you can be certain that you will draw an even sum. If you consider 11, the following possibility exists;
1+2+4+6+8+10+12+14+16+18+20=111 and that is 11 terms that sum to an odd. Any odd could work in place of the 1.
I hope this helps clarify my reasoning.
Thanks,
Jared