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# Interesting Question

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27 Jul 2004, 17:56
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Any idea how to go about solving this question ?

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?

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Senior Manager
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27 Jul 2004, 21:19
I think it is 12 cards.

The way I have thought is as follows.

1-20 have 10 even and 10 odd numbers.

The sum is a guranteed even after n number of cards. So, if we can make the sum continuously odd till a particular numbered drawn card, anycard picked up there after must make the sum even.

We know that odd + even = odd always. Thus, I pick up an odd number first - the sum is odd. An even number next - the sum is still odd. An odd number next makes the sum even. But, we are still left with many even numbers less than 20 and an even number selected makes the sum odd. Thus, even numbered sum cannot be guaranteed because we are still left with 9 even numbers.

Thus, first card is odd, the next 10 cards are even numbers less than 20. Thus, the sum is still odd. However, the next number i.e.,12th card has to be odd because the left-overs are odd only. Now, odd+odd = even.

Thus, 12th card guarantees that the sum is even.

Do let me know the OA with explanations.

ywilfred wrote:
Any idea how to go about solving this question ?

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?

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Thnx & Rgds,
Chandra

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27 Jul 2004, 23:15
Ya one another vote for 12.

as we know:
odd + odd = even
even + even = even
but, odd + even = odd ( or vice versa, even + odd = odd)

So, if we start from odd / even, till the time we pick of the same type of a number we remain at even sum but if whenever change the course we will receive ODD number
and to back on the track we need another ODD number, right ?

Lets consider worst scenerio:

we started with ODD number but next number is EVEN, we will end up with a result of ODD number. If next number must be EVEN to have summation of ODD number:
Thatway, we will end up consuming all even numbers at minimum of 11 cards already PICKED up ( 10 cards EVEN and 1 card ODD)
the twelve card has to be ODD and it will gurantees to have scene of

ODD + ODD = EVEN

Wonderful Question

Dharmin
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Perseverance, Hard Work and self confidence

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27 Jul 2004, 23:55
The OA is 12. I got the number on my second try. On my first pass, I was duped by the word 'minimum number of cards' and came up with 2, on the basis by drawing 2 even or 2 odd cards, we can get a even sum. But I guess the key word is guaranteed.

Thanks for the help !

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28 Jul 2004, 12:03
I don't agree. I think the answer should be 20.

If I chose twelve cards, and the order was:
o,e,o,o,e,e,o,o,e,e,o,o, the sume would be odd.

In order to have an even number, we have to have an even number of odd cards. But every number of cards we could pull out of the box could be made up of an even number of odds or an odd number of odds.

In the example I just wrote, there were 7 odds and 5 evens, so the answer was odd.

Just because we pass 10 cards doesn't mean we guarantee that we've gotten all the odds out of the way.

All the way up to 19 cards, we can have an odd number of odd cards. The only way to be absolutely sure that we've gotten all the 10 odd cards, ensuring an even number of odds, is to take all 20.

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28 Jul 2004, 21:47
Hi Ian,

In the example you have given, you already made the sum even at the third number itself. Following your example, the sum of the first 3 i.e., the sum of o,e,o = e.

The question asked is "What is the minimum number of cards drawn that will give a guaranteed even number ?"

Thus, the example is good for just 3 numbers. If I were to paraphrase the query given above for simplicity, I would keep it something like this.

"What is the worst case scenario where any card I pickup after certain number of cards drawn, will ensure me an even sum of all the cards drawn? In that scenario, what is the total number of cards?"

The answer does not say that it is just 12 cards picked up in any order. Rather, it is the worst case order of 1 odd follollwed by 10 even numbers(not even once the sum is even here) and then any card picked up out of the remaining 9 cards that are odd(now, the sum is even).

Thus, the answer has to be 12. Do let me know if I am not clear.

ian7777 wrote:
I don't agree. I think the answer should be 20.

If I chose twelve cards, and the order was:
o,e,o,o,e,e,o,o,e,e,o,o, the sume would be odd.

In order to have an even number, we have to have an even number of odd cards. But every number of cards we could pull out of the box could be made up of an even number of odds or an odd number of odds.

In the example I just wrote, there were 7 odds and 5 evens, so the answer was odd.

Just because we pass 10 cards doesn't mean we guarantee that we've gotten all the odds out of the way.

All the way up to 19 cards, we can have an odd number of odd cards. The only way to be absolutely sure that we've gotten all the 10 odd cards, ensuring an even number of odds, is to take all 20.

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Chandra

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29 Jul 2004, 08:00
But by the twelfth number in my list, the sum was odd. How is it that you can say that by 12 numbers I am guaranteed an even?

what if I rearranged my list to start with o,e,e... but I still had 7 odds and 5 evens? I'd have an odd number after the third and an odd number after the 12th, even though somewhere down the road I'd have an even number.

I read the question as, "how many cards to I have to pull out to be absolutely positive that under all conditions thereafter I will definately have an even sum of the numbers in my pile?" And at every possible number from 1-19, I can come up with some combination of evens and odds to get an odd number, and they are all legitimate arrangements.

A student of mine just got here. If it's not clear what I wrote, I'll complete it in a few hours....

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29 Jul 2004, 09:32
Your understanding is "how many cards to I have to pull out to be absolutely positive that under all conditions thereafter I will definately have an even sum of the numbers in my pile? ".

However, there is nothing like "we get an even number continuously thereafter". You get an even sum at any cost if 12 cards are picked up in the worst scenario as specified in the earlier mail.

The question stem is "What is the minimum number of cards drawn that will give a guaranteed even number for the sum of all the cards drawn?"

But not "an even sum thereafter". I hope I am clear now.

ian7777 wrote:
But by the twelfth number in my list, the sum was odd. How is it that you can say that by 12 numbers I am guaranteed an even?

what if I rearranged my list to start with o,e,e... but I still had 7 odds and 5 evens? I'd have an odd number after the third and an odd number after the 12th, even though somewhere down the road I'd have an even number.

I read the question as, "how many cards to I have to pull out to be absolutely positive that under all conditions thereafter I will definately have an even sum of the numbers in my pile?" And at every possible number from 1-19, I can come up with some combination of evens and odds to get an odd number, and they are all legitimate arrangements.

A student of mine just got here. If it's not clear what I wrote, I'll complete it in a few hours....

_________________

Awaiting response,

Thnx & Rgds,
Chandra

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29 Jul 2004, 10:14
mallelac,

I respect your opinion very much and appreciate this conversation. But I still don't see your point.

Even if you don't think about "thereafter" I have a way of getting an odd sum with 12 cards. So what is the "worst case scenario"? It's arbitrary.

The question wants a gaurantee of an even sum. Would you put down \$100 that if you gave me those 20 cards, and I pulled out 12 of them, I would get an even sum? I wouldn't, because I know I can get an odd sum.

I have looked for a similar question to show you why the way I'm reading it is right, but for now I can't find it. But ask yourself that question: How many cards would I have to pull out to be positive that the sum of those cards is even? How much money am I willing to bet that I'm right.

If you want a sure bet, if you want to know absolutely that the sum of the cards you pull out is even, you have to pull out all 20. Until then, there is a way to get an odd sum with every number of cards. I wouldn't put any money down on that.

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01 Aug 2004, 21:57
ian7777 wrote:
mallelac,

The question wants a gaurantee of an even sum. Would you put down \$100 that if you gave me those 20 cards, and I pulled out 12 of them, I would get an even sum? I wouldn't, because I know I can get an odd sum.

Ian,

Sure, I would put a 100\$ and gaurantee that if you keep taking cards one by one - you would get an even sum - either in a minimum of 1 card (which happens to be e) or at a maximum of 12th card (o+e+e+e+e+e+e+e+e+e+e+o)

They are asking for a minimum gaurantee.

If it is your lucky day, you pick a card and that happens to be even, say 2, and you get a even in the first card itself.

If it is your worst day, your first card is 1 (Sum is not even), second card is 2 (Sum is 3 which is not even), third card is 4 (Sum is 7 which is not even), fourth card is 6 (Sum is 13 which is not even), fifth card is 8 (Sum is 21 which is not even)..., eleventh card is 20 (sum is 111 which is not even), twelth card is 3 (there is no more even card to fool around) (sum is 114, which is even)
So minimum gaurantee is 12th card...

(I just used numbers in sequence just to be clear, the sequnece may be even 3, 4, 6, 8, 12, 10 , 16, 14, 18, 20, 2, 17)

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01 Aug 2004, 23:48
sorry, but you just lost \$100.

because I pulled out the following cards, randomly:

1, 4, 6, 8, 12, 7, 2, 10, 14, 16, 18, 17

So the 20 got left behind.

Let me pose the question in another way:

There are 20 cards, numbered 1 - 20. If a man is going to pull out 12 cards, what is the probability that the first card will be odd, the next ten will all be even, and the 12th card will be odd?

I'd work that out and say the probability is:

10/20 x 10/19 x 9/18 x 8/17 x 7/16 x 6/15 x 5/14 x 4/13 x 3/12 x 2/11 x 1/10 x 9/9 = 1/16,796

So there is a 1 in 16,796 chance that the arrangement you are guaranteeing will actually happen. I think that's not such a strong guarantee. It certainly doesn't seem to be a minimum guarantee.

Does anyone else want to suggest something here?

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02 Aug 2004, 08:33
Ian:

I admire your stubbornness. I'd have given up if someone persists so much.

My gaurantee: If you keep taking cards one by one - you would get an even sum - either in a minimum of 1 card or at a maximum of 12th card.

In other words, enroute to picking 12 cards, I gaurantee you would hit a even sum somewhere in the middle. If not, you will hit the even sum atleast in the 12th card.

Considering your example - 1, 4, 6, 8, 12, 7, 2, 10, 14, 16, 18, 17
You seem to have acheived the even sum in the sixth card itself. (We will stop picking cards once we acheive our goal of even sum).

1, 4, 6, 8, 12, 7
First card = 1, Sum = 1, Status = Failure
First card = 4, Sum = 5, Status = Failure
First card = 6, Sum = 11, Status = Failure
First card = 8, Sum = 19, Status = Failure
First card = 12, Sum = 27, Status = Failure
First card = 7, Sum = 34, Status = Success
Stop picking cards.

Another point is, this problem is number theory and not pb.
Its my goal this week to explain this and snatch that \$100...

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02 Aug 2004, 13:08
OK. I'm finally seeing what the problem is. We are reading two different questions. Thank you, hardworker_indian, for the additional insight. I now understand.

The question itself states: "What is the minimum number of cards drawn that will give a guaranteed even number for the sum of all the cards drawn?"

I read that to mean, "how many cards must be drawn to ensure that the sum of those cards is an even number?" After all, it does say "for the sum of all the cards drawn". To me, that means every card drawn. That means that if I draw 12 cards, they all must sum up to an even number

The question you are reading says, "how many cards must be drawn to ensure that an even sum has occured sometime in the past?" To you, if I've drawn 12 cards, it is absolutely true that at some point along the way, an even sum has been achieved, even if now we have an odd number.

Personally, I don't think the question asks that.

I've finally found the similar question I was looking for. Please check this out and pay attention to the wording:

Of the science books in a certain supply room, 50 are on botany, 65 are on zoology, 90 are on physics, 50 are on geology, and 110 are on chemistry. If science books are removed randomly from the supply room, how many must be removed to ensure that 80 of the books removed are on the same science?

A) 81
B) 159
C) 166
D) 285
E) 324

I will post this as a new topic for the board, but we can discuss it here, as well. Notice how detailed the question is? It specifically states that we must have 80 books in the pile of those removed on the same science before we have acheived our goal. This question cannot be misinterpreted. Please, take a shot at finding the answer. This is, by the way, a real GMAT question, word for word.

So there we go. I hope by now we all see each other's points. I see yours, but I still think the correct answer is 20, because I do not read the problem the same way you do. In my mind, I won the \$100.

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02 Aug 2004, 22:42
And in my mind, I scored 800 . I wish

Yes, I see a pattern in both these questions. And the solution too seems to be in pattern = pick all worst case scenarios and wait for the last piece to fall in place, to get a success.

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02 Aug 2004, 22:42
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