newton9 wrote:Jerome wrote each of the integers 1 through 20, inclusive, on a separate index card. He placed the cards in a box, and 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 cards he drew was even, how many cards did Jerome have to draw?

Answer is 12

there are 10 odd numbers and 10 evens with us.

Also, we know that Odd + Even = Odd is the only possibility which will yield Odd as the sum.

Aim : try to find out the extent till which you can get odd as the result.

For calculation purpose, you can start with either an even number or an odd.

Suppose, the first one Jerome picks is an odd number. Then he gets an even, then even, then even ........ he exhausts all the evens. But the sum is still Odd. Once he's done with all the evens, he'll have used up 11 numbers. The twelfth will be an odd, which when added to the sum will SURELY give us even result.

Hence, the min. number to have even as the sum = 12.

This is an ambiguous question that often leads to confusion. See also:
www.beatthegmat.com/hard-number-problem ... t8767.html
www.beatthegmat.com/pr-practice-test-problem-t1664.html

clickclickdecker wrote:This is an ambiguous question that often leads to confusion. See also:
www.beatthegmat.com/hard-number-problem ... t8767.html
www.beatthegmat.com/pr-practice-test-problem-t1664.html

yes, the language is a bit ambiguous. Anyway, there's an award for me . The first thing/interpretation of Stuart was my first interpretation as well !!!! yippieeeeee!!!!

As far as the question is concerned, it can be made unambiguous by saying that "What is the minimum number of cards that can be drawn to ensure that the sum is odd? "