If we want to figure out the number of cards he needs to draw to guarantee even, let's see how long a string of odd sums we can make.

We have 10 odds and 10 evens.

If we start with even, we're done.. so we must start with odd.

If we ever draw an odd again, we'll have an even sum (O+O = E). So, to keep our sum odd, we need to keep adding even numbers.

So, the longest string we can make is:

odd + (our 10 evens).

After 11 cards we're out of even numbers. So, the 12th card will guarantee an even sum.

thanks Stuart.

I think my thinking process/approach is wrong as i did not arrive at the correct answer, but cannot figure out where ?

This was my take on it....
So if we say that answer is 12. Jerome needs to pull out 12 cards to gurantee that the sum is even.
Now what if he pulls out 7 even cards and 5 Odd cards (7 + 5=12 cards)
sum of 7 even cards = even
sum of 5 odd cards = odd

and finally even + odd = odd. (final sum is odd!)

How does 12 cards gurantee that sum WILL be even?

rainbownlife wrote:thanks Stuart.

I think my thinking process/approach is wrong as i did not arrive at the correct answer, but cannot figure out where ?

This was my take on it....
So if we say that answer is 12. Jerome needs to pull out 12 cards to gurantee that the sum is even.
Now what if he pulls out 7 even cards and 5 Odd cards (7 + 5=12 cards)
sum of 7 even cards = even
sum of 5 odd cards = odd

and finally even + odd = odd. (final sum is odd!)

How does 12 cards gurantee that sum WILL be even?

Hmm.. maybe I misinterpreted the question. I interpreted it as "to ensure that along the way he'd get an even sum", rather than "at that number the sum will definitely be even".

Is 12 the right answer? If it is, then I interpreted it correctly. If it isn't, then the only number of cards that will guarantee an even sum is all 20. For every other number we could have either an odd or even sum.

Hi Stuart
The correct answer is 12. Your interpretation is correct.

Q asks:
In order to ensure that the sum of all cards he drew was even, how many cards did he have to draw?

Can you elaborate on why it is interpretation 1 instead of 2?

interpretation 1:
"to ensure that along the way he'd get an even sum"

interpretation 2:
"at that number the sum will definitely be even".

I'm actually not sure why that 1st interpretation is correct, it was just the first thing that came to my mind when I read the question. Given that both interpretations are valid, this question wouldn't make it to the actual GMAT.

The question indirectly asks for the worst case in which you get a sum of numbers to be even (on the whole not along the way).
The answer 12 is correct !
It cant be 7 even and 5 odd because in the case you hit an even sum at the second instance of picking an even numbered card!...

Hope its clear to an extent

Thanks
Senthil

Hi Senthil
not completely clear...but may be kinda....will have to think more abt it

Hi Stuart
You definitely got the right answer first in your mind for this question ....

Stuart Kovinsky wrote:I'm actually not sure why that 1st interpretation is correct, it was just the first thing that came to my mind when I read the question. Given that both interpretations are valid, this question wouldn't make it to the actual GMAT.

Stuart, I fully agree that this question couldn't make it to the GMAT (or if so, they'd have to adjust some scores once someone proved this wrong ). I also, challenge you in your original logic (though your logic is exactly that of the author of this question in Princeton Review). Here I go...

So you say that we must pick 12 numbers. Well, how about this set of 12 [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14]? Lets take a look at the sum of the first 10 numbers. We get 1+2+3+4+5+6+7+8+9+10 = 55, so we have an odd number.

Well, we have 5 more even numbers to choose from now [12, 14, 16, 18, 20]. If two of these are selected in your final two (as I chose 12 and 14 as my final numbers), you get an odd sum. In my scenario, the sum is 71.

Did I miss something, or shall we contact the writers of Princeton Review and tell them they are wrong!?!

Ian Stewart wrote:

tarn151 wrote: So you say that we must pick 12 numbers. Well, how about this set of 12 [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14]? Lets take a look at the sum of the first 10 numbers. We get 1+2+3+4+5+6+7+8+9+10 = 55, so we have an odd number.

Yes, those numbers give you an odd sum, but if you stop picking numbers the instant your sum is even, you'll never pick those ten numbers. If your first selection was even, you'd stop. If you're first selection was odd, then you must keep picking even numbers in order for your sum to remain odd. So after you pick ten numbers, if your sum is going to be odd, you have to have one odd number and nine even numbers - you can't have the set of numbers you list above.

I'd add that I really dislike the wording of the original question, which I find imprecise. You wouldn't see a question worded in such a way on the real test.

True, but the question doesn't say that he'll stop once he gets an even sum or that he adds them up as he goes. So it could just be that he selects a group of cards face down and then adds them up once he's all done. Either way, a very poorly worded question.

tarn151 wrote: True, but the question doesn't say that he'll stop once he gets an even sum or that he adds them up as he goes. So it could just be that he selects a group of cards face down and then adds them up once he's all done. Either way, a very poorly worded question.

Which is exactly what 3 other people said in the thread earlier on (including me)!

Thanks for this discussion people. This question has literally been driving me nuts. On a test (even a practice test) where wording seems to be so important, a question like this can really mess with you.

He got 20 cards in a box, then he picks cards "randomly". How does he ensure he draws an even sum? Really, he could draw anything, including 3 odds and 9 evens which would give him 12 cards with an odd sum. No matter how hard I try, I fail to see the part about him stopping when he gets an even sum.

This thread is pretty much resolved but I felt like sharing my frustration, I guess lol.