Problem Solving

Jerome wrote each of the integers 1 through 20, inclusive, on a separate index card. He placed the cards in a box, then drew cards one at a time randomly from the box, without returning the cards he had already drawn to the box. 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 11
D 10
E 3


Source: Princeton Review

Someone able to explain?

IMO. Ans is B

sreak1089 wrote:IMO. Ans is B

thats correct, mate.

how did u get it?

I don't think the question is right.
If the answer is B.
let's say we select 9 even numbers from 2 to 18
and odd numbers, 1,3,5
the sum is odd.

same for all the options.

Figure out how many ways the sum would remain odd...

First pick an odd number, say 1
Then you can pick every even number and add it to 1, and the sum will still be odd.

1+2+4+6+8+10+12+14+16+18+20= 111

Now all you have left to pick from are odd numbers, which will make the sum even.

111+3=114

So it took a maximum of 12 numbers to insure the sum would be even.

Liquidsilver wrote:Figure out how many ways the sum would remain odd...

First pick an odd number, say 1
Then you can pick every even number and add it to 1, and the sum will still be odd.

1+2+4+6+8+10+12+14+16+18+20= 111

Now all you have left to pick from are odd numbers, which will make the sum even.

111+3=114

So it took a maximum of 12 numbers to insure the sum would be even.

like I said in my previous post..'
what if i select 9 even numbers and 3 odd numbers

oh this question is retarded. i didnt think that jerome would see the cards that he draws. i always thought it was like him drawing the cards not knowing what he drew.... but if he sees the cards than this question is just plain stupid.

thanks guys

In order to guarantee that the sum is even, we need to think of the worst case scenarios as follows:

scenario 1) odd + even + even + even ...... + odd.. --> at most there are 10 even numbs hence 10 evens + 2 odds = 12 cards to guarantee an even sum

scenario 2) even + odd + odd + odd.....+ even...-->10odds + 2 evens = 12 cards to guarantee an even sum

both of the above guarantee an even sum in the longest and surest way.

hope that helps.

The way I approached was to eliminate answer choices which would not ensure an even sum and ended up with 12. But then on a second thought, I thought 12 will not be an answer because, say you picked 3 odd numbers, 7 even numbers initially and the next 3 happen to be odd numbers, then 12 won't ensure that the sum is even.

I suspect this question is wrongly worded, IMHO.

sreak1089 wrote:The way I approached was to eliminate answer choices which would not ensure an even sum and ended up with 12. But then on a second thought, I thought 12 will not be an answer because, say you picked 3 odd numbers, 7 even numbers initially and the next 3 happen to be odd numbers, then 12 won't ensure that the sum is even.

I suspect this question is wrongly worded, IMHO.

In your example you would stop once you had pulled the second odd number, since your sum would then be even. There would no reason to continue pulling new cards. The use of the word "ensure" is what I think some people is missing. While you can pull fewer than 12 to attain a sum that is even, you can never pull more than 11 and obtain an odd sum. So that 12th card ensures the sum will always be even.