The odd ball

Use the "half-half" method (my terminology).

1. With twelve balls, put six balls in each pan. The pan that is heavier contains the odd-ball. Set aside the balls in the lighter pan.
2. From the remaining six balls, put three balls in each pan. The heavier pan contains the odd-ball. Set aside the three balls in the ligher pan.
3. From the remaining three balls (one of which is the odd-ball), put one ball in each pan. If the pan tips, the odd-ball is the one on the heavy side. If the two balls weigh the same, the odd-ball is the one that you haven't weighed.

As to the question of where to go...

If I had a wife like yours, I'd ask her where she'd want to go. A wife like that is worth more than a golf holiday.

^{I can almost hear the diehard golfers out there...YOU'RE GOING TO SAY WHAT???!!!}

Vulcan,

My wife thanks you, but, wait a minute. I never said the odd ball was heavier. It could be lighter. Weighing six against six, you are sure to learn that one pan is heavier, but if the odd ball is lighter, you still don't know any more than when you started.

Lift it, turn your head, and cough!

Good answer, Verm.

I think Vulcan's answer is the only answer which is also correct. It does nor matter whether one ball is heavier or lighter, three weigh attempts is clearly showing the odd ball. Others may sretch this thread but I think it is solved.

Now where you go for golfing, ask Ma'am to tell you where she wantes to go and while she is shopping, you may putt around within that time frame.

By the way what did she give you this Christmas as you have not utilised her first offer yet?

I hope you both along with the family had good Christmas. Best of luck for New Year and beyond.

Nadeem,

Your powers of deduction are amazing!

In response to your question, she gave me another dozen golf balls because I lost the ones I got last year, just like Del. All twelve of the new balls are identical in weight, mass, volume and density.

Happy New Year to all.

I still think my answer is correct... #28

I liked your answer in all respects, as I am sure your wife would too!

Yes, she did. As I read ba/el's reply, she was reading over my shoulder .

Unfortunately, my answer wasn't correct and I had to go back to the drawing board. Interruptions and things to do got in the way and by the time I figured it out, the thread had grown in my absence and others had already submitted the correct answer.

Now, my wife's asking for a vacation too.

A truly great reply on all points. Your wife should also be proud.

I thought you were the #1 in responding the thread with the correct answer. I do not see any thing wrong with your anwer. Please tell me how your answer is wrong? ( That was my answer too and you beat me to write it and I am defending it).

Hi Nadeem430,

Read ba/el's response in post#2.

I assumed the odd-ball was heavier. The question doesn't say that. It can be heavier or lighter. That complicates the problem by several folds.

This was a great exercise! I was sharing it with my 13 year old daughter who's just learning algebra. I didn't use algebra (trial and error's the best I could do) but she had a go at it. She, too, assumed that the odd-ball was heavier and got the same answer that I did. I was real proud of her.

Hi Vulcan:

It does not matter whether it was heavier or lighter. All we are doing is the separate out the group which is not equal and then for third weighing, it will give the correct odd ball. I shall stick to this answer.

I am also proud of that little angel sitting with a be--, no, the Vu-- and came up with the correct answer.

Have a good day and enjoy it.

I was going for 4 - 4 , 2 -2 and then 1- 1 but not knowing what to look for it is back to the drawing board for me.

May I bounce the balls?

Of Course it can be done.

And you are welcome to Shillong/NE India

Shall be glad to welcome you and your witty Good Lady!

Bruce your second measurement of 3 known and 3 unknown will determine if oddball is light or heavy.

I would suggest Traverse City, Mi. in late summer, a dozen top-notch courses in the area with no crowds.

Thank you all for your help. I think ConcernedConch has come up with the answer. If you haven't followed it, let's start by numbering the balls...then:

First Weighing: 1,2,3,4 vs 5,6,7,8

If they balance, odd ball in 9,10,11,12; go to a1

else Second Weighing: 1,2,5 vs 3,4,6 (3,4,5 have switched pans)

if they balance, odd ball in 7 or 8; go to a2

else if pans tip same way in First and Second weighing, odd ball in 1,2 or 6; go to a3

else odd ball in 3,4 or 5; go to a4

a1: Second Weighing: 9,10 vs 11,1

if they balance, odd ball is 12

else Third Weighing: 9 vs 10...etc.

a2: Third Weighing: 8 vs 7. If pans tip same as First Weighing, odd ball is 7 else odd ball is 8.

a3: Third Weighing: 1 vs 2...etc

a4: Third Weighing: 3 vs 4...etc.

Thanks for a couple of great venues for my golf holiday.

My head hurts. Is there a formulism to attack logic problems like this? I'd love it if someone clued me in.

Check out the hyperlink in post #14. Then look at post #20 within that reference. Your head will feel like a million dollars.

But at this point you only know that one group is different. If the ball is heavier (or lighter), it will be obvious - but you need two more weighings to find the different ball. Only a fourth weighing will differentiate the odd ball.

Hit a few holes in one for me, ba/el.

I tried once, got my new special golf bag down to the golf links, set my golf ball onto the tee, prepared to swing and that, as you can see from the pic, was it.

I never got to scale the dizzy heights of trying to weigh the individual golf balls.....

That is very simple. First weighing, you weight 6 balls on 2 sides. You will find the heavier group. Then, weight the heavier group with 3 balls on both sides. You will find the heavier 3 balls group. Then, weight 1 ball on 2 sides, if they are equal, the one which is not weighted is the heavier one. I will say Golfing in Pebble Beach California is your best option. Please see site link from here http://www.pebblebeach.com/page.asp?id=704.

Tai

Hi,

Surely you can do it. First put six balls on each pan. Then the group having a denser ball will go down. Eliminate the other group.

Now put three balls each on pan, and again eliminate lighter group. Now you will be left with three balls, in which one is denser; and one weighing. Take any two balls. If they balance, the ball not put on pan is the odd ball. If they do not balance, the answer is obvious.

And where should you go to play golf?

Sorry, I don't play golf.

Hirematth

i think if you way 6 against 6 taking the lightest 6 then 3 and 3 taking the lightest 3 then 1 and 1 leaving 1 ball out it would be obvoius which ball is the lightest.

Wasn't there a challenge with gold coins that was basically the same thing you have encountered? Check the old challenge questions. the answer may be already there.

http://cr4.globalspec.com/blogentry/983/Counterfeit-Coins-Newsletter-Challenge-01-23-07

Try above link.

http://cr4.globalspec.com/blogentry/983/Counterfeit-Coins-Newsletter-Challenge-01-23-07

check here. Might help. Sorry for the repeat. I didn't think the link took the first time.

Charsley99 is absolutely right! The problem is the same as the counterfeit coin. Sorry about that. I guess it goes to show there's nothing new under the sun.

Merry Christmas, everyone.

I would'v done that!

KISS

I'm really amazed, I can't believe such a collection of well brained people don´t know this old problem.

1. You must divide the 12 balls (or coins as I learned when I was a child 50 years ago) into three equal parts, say balls nºs 1,2,3and 4 (group 1), balls 5,6 7 an 8 (group 2) and last group 3 (balls 9, 10, 11 and 12).
2. Choose any two groups and put them one in each balance side. At this point two things may happen:
1. The weight is balanced: all 8 balls used are good, and the odd ball must be in the third group not weighed.
2. The balance swings: in this case the odd ball is in the balance and NOTE that if the odd ball were in the upper balance arm, It has to be lighter than good ones. If the odd ball were in the lower arm, obviously it has to be louder than good ones.
3. Second weight if case 1 happen: Take 3 of the remaining balls and put them in one arm, leaving 3 good balls in the other arm. Again two thigs coud happen:
1. Weights are balanced: in this case the odd ball is the fourth one still not weighed. Just put it in an arm against one good ball in the other. I think all of you can in this case know by the balance position if the odd ball weights more or less than good ones. (Success in 3 weighings)
2. Weights are unbalanced: The direction of fulcrum swinging will tell you if odd ball (one among the three weighed) is lighter or louder than goods ones. The third weight in this case must be to put two of that balls one on each side of the balance. NOTE that you Know in this case the type of odd ball (lighter or not). If the two balls are balanced, the odd is the remaining one. If fulcrum swings the ball which goes in the correct direction is the odd one. (In this case SUCCESS again in 3 weighings)
4. Second weight in case 2 above. Tag the two groups as lighters (those who went up) or heaviers (those who went down). Put on one balance arm 3 "lighter" and one "heavier" ball and the remaining "lighter" wiht 3 good balls in the other arm. YES, you're right: again two cases may happen.
1. If weight is balanced the odd ball must be among the 3 "heavier" not included in the last weighing. Weight two of them one against other and if balanced the odd is the third . Otherwise the ball in the lower position is the odd (remember are tagged "heavier") AGAIN SUCCES in 3 weighings.
2. If weight is unbalanced the odd ball is obviouly present. If arm with 3 "lighter" and one "heavier" is up. The odd ball must be one of the 3 "lighter" to distinguish between them the third weighing as previous cases. NOTE that the fourth "lighter" in the other arm must be good: if it's lighter, the arm in which is located cannot go down!!!!. If the mentioned arm with 3 "lighter" and one "heavier" is down, the odd ball can be either the "heavier" in this arm or the "lighter" in the other arm. As both are tagged, take anyone of them and check it against one good ball and ... BINGO. The problem completely solved.

I would half the 12 balls in 2 trays ,pick the heaviest tray and half the balls again the three balls that are left I would put one in each tray if they equal out then the odd ball is in my hand if not the odd ball is in the heavier tray.

Sorry, guest. No golf holiday for you. But Merry Christmas anyway.

I was just commenting elsewhere that the number of 'homework problem' posts had slowed down with the kids out on holiday -

Is this not actually a contrived way of presenting a non-problem and shouldn't there be a rating system for homework misdirection and old riddles?

If I repeat the same thing that has been repeated 12 times do I get a prize?

"If I repeat the same thing that has been repeated 12 times do I get a prize?"

Yes, The Awards Committee may decide to give you this

Now you know...Happy Christmas....

1. Number the balls 1 to 12.

2. Weigh balls 1,2 & 3 against 4,5 & 6. --> First weighing

3. Balanced? Continue to step 4. If not go to step 9.

4. Weigh 7, 8 & 9 against 1, 2 & 3. Balanced? Continue to step 5. --> Second weighing. If notbalanced, we know that the odd one is one of 7,8 & 9 and also whether the odd one is lghter or heavier. Go to step 8.

5. The odd ball is one of 10, 11 &12. Weigh 10 against 11. Balanced? Continue to step 6. If not go to step 7. --> Third weighing.

6. The odd ball is No. 12 and found in third weighing..

7. Bad luck! You need a fourth weighing.

8. Weigh 7 against 8. Balanced? Then the odd one is No. 9 If not the odd one is the lighter or heavier of 7 & 8 as determined in the second weighing. --> Third weighing and the answer is found.

9. Note whether 1,2 & 3 are lighter compared to 4, 5 &6. Weigh 7,8 & 9 against 1,2 & 3. Balanced? Then go to step 10. If not go to step 11. --> Second weighing.

10. The odd one is one of 4, 5 & 6 and also we know whether the odd one is heavier or lighter! Weigh 4 against 5. If balanced the odd one is 6 and if not the lighter or heavier one of 4 & 5 as noted. Answer found! --> Third weighing.

11. The odd one is one of 1,2 & 3. Weigh 1 against 2. If balanced the odd one is 3. If not, the odd one is the lighter or heavier of 1 & 2 as noted. Third weighing and the answer is found!

Dear ba/ael, you have a very good chance (11 out of 12) of winning the golf vacation promised by your wife! I suggest that you take your wife on a vacation that she would like to have!

Dear Ulrasetaknev,

I have an even better chance than 11 out of 12, partially because my first weighing would be four against four rather than your suggested three against three (see posts #7 and #14) and partially because my wife is an even more enthusiastic golfer than I am. Not likely to be too much resistance there. So, in effect, I am taking your advice and providing her with the kind of vacation she would like to have. Thank you and a very merry Christmas.

Perhaps we should terminate this thread here and now. I wasn't aware that it had been discussed in an earlier issue of GlobalSpec. Some of the CR4 group are finding it a bit boring. Can't have that, can we?

The ones that are bored can leave, so don't worry about that, but i also feel that we have "played" it out too, so bye and have a great 2008!

Hi BA/AEL,

You still have not provided a gauranteed solution. The coating on a golf ball is only a fraction of an inch deep. The filling inside the ball could be substituted (you said nothing about this) for another type of inside material with a different weight therefore it would be possible that even with a diiferent density the weight of the ball could be the same as the others and the scales would balance regardless of which balls were being weighed. The outside of the golf balls could still look the same.

yes it can be done ! 1st weight ,6 and 6,get the 6 heavier ones.

2nd weight,3 and 3, get the 3 heavier ones.

3rd weight ,weight any two balls,if the balance equalizes the third ball is the heavier one if not you got your heavier ball in the balance.

Go to Aruba for your golf Holyday!!!!!!

nadeem has the right answer, come to fla.

it was really amazing...

hahahahha anybody with another this kind of question to share? i really feel the blood flowing in my head. but it feels good...

There are a lot involving astrophysics and cosmology. Check some of them out for a real mind bender!