Princeton Review Cracking the GMAT Practice Test Question

[abramson]

Jerome wrote each of the integers 1 through 20, inclusive, on a separate index card. He placed the cards in a box, then drew one card at a time randomly from the box, without replacing the cards he had already drawn. In order to ensure that the sum of all the cards he drew was even, how many cards did Jerome have to draw?

A. 19
B. 12
C. 13
D. 10
E. 3

_________________________________________

The book says (B), but I think this is flawed - none of these could be correct answers. Does anyone agree/disagree? Thanks for explaining your logic.

[anadi]

Worst case is that first draw is odd, and second onwards everything even till all even are over. So 1 odd + 10 Even + 1 more odd to make the sum even. Total 12.

[Harish Dorai]

I honestly didn't understand the question. What does the statement "sum of all the cards" mean? The word "ALL" is ambiguos.
From 1 to 20, we have 10 odd and 10 even numbers. So if ALL means all the 20 cards, then without drawing any card, I can tell that the sum will be Even. So the answer is 0 ;-)

[abramson]

The question asks how many cards he needs to pick out at the least, to ensure that the sum of the numbers on all the cards he has picked out is even.

anadi - The book uses a similar "worst case" scenario and says if you picked out an odd first, then even if you picked out all 10 even cards, you would have 11 cards, summing to an odd number. Then, since you would not have any even numbered cards left, the next card (12th one you pick) would be off, making the sum now even.

However, my doubt it this: What if the 12 cards you picked out comprised of 7 odd and 5 even cards (and I don't know why this couldn't be possible since you are picking out at random) - then you would still have an odd sum. So how does he 'ensure' that he has an even numbered sum with 12 cards?

I might be missing something basic here I feel. Thanks for any help!

[ad]

The case that anadi points out is the worst case. We can't assume things like 7 odd 5 even. If that be the case we could have got an even sum much earlier. This gives the worst situation when you keep on drawing cards till you get the even sum.

[esledge]

abramson, you are absolutely right on this one--the answer cannot be 12. In fact, I think the correct answer would have to be 20.

In cards numbered 1 through 20, there are 10 Evens and 10 Odds.

Drawing 12 cards, you can have a variety of possibilities:
10 Evens + 2 Odds = Even sum
9 Evens + 3 Odds = Odd sum
(and so on...)
Already we can see that selecting 12 cards doesn't guarantee an even sum.

Even selecting 19 cards wouldn't do it:
10 Evens + 9 Odds = Odd sum
9 Evens + 10 Odds = Even sum

But 20 cards would, as there is only one way to select every card in the set:
10 Evens + 10 Odds = Even sum

It sounds like the published solution made the same mistake as other posters here. Beware of thinking of the "worst case scenario" as a string of Evens or Odds. Here, I suppose the worst case scenario is any sum that ends up being odd, and the order of selection doesn't matter. It's better to try to produce an odd case and an even case for the number of cards selected; it should be fairly easy to come up with one example of each if such an example exists.