David wrote each of the integers 1 through 20, inclusive, on a separat

Thanks TheDoberman !!

What is chance that all cards he draws is even only? In that case why cant he draw 10 even and still ahve sum even? This is where I am confused?
My answer was since he can draw 10 even or odd cards that gives an even sum 10 is answer!

Can you give a scenario as you quoted above of answer less than 12 will end up odd...perhaps that will clarify things a bit.

Thanks in advance.

The question rephrased should be what is the most cards that he could draw before it has to sum to an even number;

In 1-20 there are ten odds and ten evens. Thus the most cards you could draw would be if the first sum is odd and then you keep adding evens until you run out then add an odd number.

odd+even+even+even+even+even+even+even+even+even+even+even=even; actual numbers
1+2+4+6+8+10+12+14+16+18+20+3=114 or
3+2+4+6+8+10+12+14+16+18+20+1=114 or other combinations of 2 odds.

It would take 12 terms.

Thanks,

Jared

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

Thanks guys! Great clarification...got it now...wording isn't great as you said in the question.

Hrmm... good question. I was also confused with the wording. I wasn't aware that he stops once he reaches an even sum. (because you can basically reach an odd sum with any combination of numbers, ie 7 Odds + 5 Evens gets you an odd)

Makes sense now, thanks for the clarification guys.

Please respond...
Can this be a valid question??? It is told RANDOMLY in the answer.... picking up all even card is absolutely not random.... I am not sure if this is a valid question

It is possible to randomly pick all the evens in a row. I may not be likely, but it is possible.

Thanks,

Jared

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

Bunuel,

12 is correct. And that is the max draws you need to definitely know that the sum of drawn cards is even.
Say the first card you draw is Even... then
E..O...O = Even sum
E...E = Even sum
E...O..E...E...E..E....9 more E cards....O....(12th card) has to odd... sum is even..

Say the first card is odd..
O...O = even
O...E..E...10 more E cards max..O... = even

I can't understand how you say 20 is the answer. Definitely before the draw of 20 cards you will have a sum which is even.

I know you said you don't care about the question ..but just academically..what do you say?

Thanks.

Please read my solution above.

The answer depends on the interpretation of the question. If you interpret it as at which drawing the sum will definitely be even then 12 is wrong, as for example you can draw 7 even numbers and 5 odd numbers so after 12 drawings you will have an odd sum. So if it is the question then the only # of drawings that will guarantee an even sum is 20, because for every other # of drawings you could have either an even or odd sum.

Again not a good question.

If you draw 7 even numbers and 5 odd numbers you will have an even sum much before 12 cards. Do you see that? Anyway..time to move on!

I don't understand what you mean by that.

But again: the question allows 2 interpretations, one gives the answer 12 and another gives the answer 20. There is no reason to believe that 1st one is better than the 2nd one. That is why this is a bad question.

You'll never see such an ambiguous question on GMAT.

Hope it's clear.

After being confused for a while, the solution seems clear to me.

The answer is B - 12
Keeping in mind that Jerome can stop in the middle when ever he has the sum as even .

Consider he starts with even , he stops . He has the sum as even .
Consider he starts with ODD, he can not stop. At this point, he either has the bad luck of drawing all even numbers there of . He would now draw the 12th card , which would be ODD ,

O+(EVEN Series)+O = E .

Do note, IMO the question should not have been "how many cards did Jerome have to draw" but rather should be "how many maximum number of cards might Jerome have to draw" .

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

This is not a good question. The answer depends on how you interpret it.

The given solution suggests that they meant the maximum number of picks needed to ensure that on the way you had an even sum. In this case worst case scenario would be to draw odd number first, and then keep drawing all 10 even numbers. After 11 draws you'd still have an odd sum and need to draw 12th card, which will be an odd to get the even sum.

But if you interpret it as at which drawing the sum will definitely be even then the answer would be 20. As only on 20th draw I can say for sure that the sum is even.

I wouldn't worry about this question at all.

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