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# There are 20 cards in a box and each card is numbered from 1 to 20. Bi

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There are 20 cards in a box and each card is numbered from 1 to 20. Bi [#permalink]
If you pull 1 then every even card in a row you'll get {1,2,4,6,8,10,12,14,16,18,20} then you'll need one more to get an even sum. That set is 11 integers, so I'd guess the answer is 12, because you'd have to pull one more card and it would be odd, since there are no evens left: hence 12 cards would be pulled and guarantee an even sum.

I venture to guess D.
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Re: There are 20 cards in a box and each card is numbered from 1 to 20. Bi [#permalink]
Bunuel wrote:
There are 20 cards in a box and each card is numbered from 1 to 20. Bill picked cards at random from the box without replacement. What is the minimum number of cards drawn that will give a guaranteed even number for the sum of all the cards drawn?

A. 1
B. 3
C. 10
D. 12
E. 20

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Re: There are 20 cards in a box and each card is numbered from 1 to 20. Bi [#permalink]
I'm confused with the wording. It's saying the minimum number of cards drawn where the sum is guaranteed to be even. The answer is saying 12 but if you did something like draw 9 evens and 3 odds that would be 9 evens so even + 3 odds so odd so basically even + odd = odd. The sum for 9 evens and 3 odds is odd not even, so how can it be guaranteed that for 12 cards drawn, the sum will be even? This wording makes it seem like 20 is the answer, as you can only be guaranteed that the sum will be even after all cards. If it was something like what's the minimum number of cards needed to guarantee that the sum of those n cards will have been even at least once, then it would make sense. In short, all cards drawn doesn't mean that no even sum ever occurred.
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There are 20 cards in a box and each card is numbered from 1 to 20. Bi [#permalink]

masondr wrote:
There are 20 cards in a box and each card is numbered from 1 to 20. Bill picked cards at random from the box without replacement. What is the minimum number of cards drawn that will give a guaranteed even number for the sum of all the cards drawn?

A. 1
B. 3
C. 10
D. 12
E. 20

I'm confused with the wording. It's saying the minimum number of cards drawn where the sum is guaranteed to be even. The answer is saying 12 but if you did something like draw 9 evens and 3 odds that would be 9 evens so even + 3 odds so odd so basically even + odd = odd. The sum for 9 evens and 3 odds is odd not even, so how can it be guaranteed that for 12 cards drawn, the sum will be even? This wording makes it seem like 20 is the answer, as you can only be guaranteed that the sum will be even after all cards. If it was something like what's the minimum number of cards needed to guarantee that the sum of those n cards will have been even at least once, then it would make sense. In short, all cards drawn doesn't mean that no even sum ever occurred.

­
Consider this scenario as a game where the game ends as soon as the cumulative sum of the cards drawn becomes even. The question asks for the minimum number of card draws necessary to ensure that the game would have ended at that point. The correct answer is 12 because this is the least number of draws required to achieve an even sum at some point during the process.

Here’s why 12 is the minimum:

• If we draw 11 cards, there's a possibility that we could select 1 odd card followed by 10 even cards, resulting in a continuously odd sum throughout these 11 draws.
• However, when we extend the draw to 12 cards, an even sum is guaranteed. For instance, if the first 11 cards include 1 odd and 10 even numbers (resulting in an odd sum), the next card drawn must be odd (since there are only 10 odd and 10 even numbers in total). This additional odd card changes the cumulative sum from odd to even.

Thus, at the 12th draw, it is mathematically impossible not to achieve an even sum at some point in the process, which ensures the game would have ended by this draw.

P.S. I agree that the question could have been worded better.­
There are 20 cards in a box and each card is numbered from 1 to 20. Bi [#permalink]
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