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".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?
....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 wrongStuart 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.
). 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...
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.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.
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.Ian Stewart wrote: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.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.
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.
Which is exactly what 3 other people said in the thread earlier on (including me)!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.
