Q3
3. 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?
posted by Ramkumar @ 9:11 PM
3 Comments:
At 8:03 AM ,
kart said...
Provide me the options da...
Have to draw 20 cards to be sure that the sum is an even number.
At 1:18 AM ,
Ramkumar said...
Lets assume the worst-case scenario.
Jerome draws an odd. Then an even. This gives an odd number. Next he draws another even. Now we have an odd again. Then he draws an even. Again the sum is odd.
So, to sum it up,
We have 10 odd and 10 evens.
If his first draw is odd, then the next 10 are even, we still have an odd sum. The tie breaker will be the 12th card, which now has to be odd since all evens have been exhausted. So if the first card drawn is odd, then we must DRAW 12 CARDS.
If the first is even, then the second is odd, again we have an odd number. Now we only have 9 evens lefts, we must exhaust all of them to get an odd one. So again, 12 cards.
So the answer is 12. The 12th draw ensures an even sum.
This is the answer given but i too think it is 20.
At 2:22 PM ,
Unknown said...
The reasoning provided is hogwash and the answer of "12" is incorrect. 0 or 20 are the correct answers.
You can construct a counter-example to the answer "12" by adding the first nine odds and the first three evens. They sum to 93. Therefore Jerome is not guaranteed an even sum after 12 draws.
He must draw all 20 cards to be absolutely sure he has drawn an even number (all 10) of the odd cards. Any less than that and he could draw an odd number of odd cards, resulting in an odd sum.
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