# 8 balls, 2 weighings

1. Sep 20, 2007

### jojo_working

The President is going to a ping-pong match in Tokyo. He is taking
8 balls with him. A terrorist has implanted one of the balls with
explosives, but they add such a small amount that no one can tell the
difference. The CIA has only enough time for TWO weighings before the
match.

I know that you must weigh more than one ball on each scale and that
you can get information from the good batch of balls. Can you help?

2. Sep 20, 2007

### Capuchin

Erm, I hate to be a pedant and all, but surely you only need a single ball for a match, so you just need to make sure that one ball has not got the explosive, you can do this by splitting the 8 balls into groups of 4, weigh 2 balls on each side for each group. The group which balances is safe giving you 4 balls to play ping-pong with, which is plenty. Dispose of the rest.

3. Sep 20, 2007

### Jimmy Snyder

Yeah, the same problem can be stated with a much better scenario. Bringing ping pong balls to Tokyo? They have ping pong balls there already that you can use. Toss the entire batch and get new ones. The CIA only has time for two weighings? I'm sorry Mr. President, but we didn't have time to check thoroughly, you'll have to take your chances.

4. Sep 20, 2007

### Capuchin

The problem doesnt even state what you're trying to achieve with these weighings? It also doesnt state whether you're using a balance or scales.

If you got this from a teacher, I would write an answer discussing the many holes in the question.

5. Sep 20, 2007

### Staff: Mentor

Don't be so picky. This is an old problem, retold. Getting rid of all of the story-line baggage,

You are given eight objects, seven of which have identical weights and the other of which weighs slightly more than any one of the seven identical objects. You are to use a balance scale and a balance scale only to identify the heavy object. Moreover, you can only use the scale twice.

If you only had two objects you would simply weight the two objects against each other. The heavier object is instantly identifiable. The eight object problem can be solved with only two weighings. The trick, as the OP noted, is to weigh more than a pair of objects at a time (at least the first time, anyhow).

Last edited: Sep 20, 2007
6. Sep 20, 2007

### Capuchin

If you know that the offending ball weighs more, then you can weigh 3 vs 3, if they balance then measure the other 2 and the heavier one is your ball, if they dont then weigh 2 of the heavier 3 against each other, if they balance it's the left over one otherwise it's the heavier one.

Is it possible if you don't know that the other ball is heavier? Or do you need 3 weighings?

7. Sep 20, 2007

### Wonderballs

You never specified which president was playing the match, you could have a variable of answers depending on your own political views.

8. Sep 20, 2007

### Capuchin

You have two cows...

9. Sep 20, 2007

### NeoDevin

Maybe they're trying to find the one with explosives in it so they can give that one to the president. Given the current president of the US, that would seem the better option.

10. Sep 20, 2007

### arildno

This is the kind of terrorist dilemma that proves you should only handle your own balls, and never anybody else's.

11. Oct 15, 2007

### Wild Angel

Weigh 2 balls if there is a mass difference discard the one with the higher mass and if they are the same take both and discard the rest or if they still aren't sure discard all .Im sure Tokyo is at no shortage of ping pong balls. by the way I agree with NeoDevin

Last edited by a moderator: Oct 15, 2007
12. Jan 4, 2008

### Werg22

Ok, it took a good 20 minutes but I've got it. Label the balls a, b, c, d, e, f, g, h. Weight a, b, c against d, e, f. If they level out, then weight g against h and you're done. If they don't level out (for simplicity we'll say that a, b, c is heavier), then weight g, b, d against a, e, f. If they level out, then it is c. If g, b, d is heavier, it is b. If a, e, f is heavier, it is a.

Edit: I overcomplicated things. It's much more simple to weight d against e for the second weighing.

Last edited: Jan 5, 2008
13. Aug 19, 2009

### pingpongballs

Assume x = bomb, o = normal

Weigh 3 random balls against 3 other random balls, so that 2 are unweighed.

If the 3 vs. 3 balances, then one of the 2 unweighed is the bomb. Weigh those two and the heavier one is the bomb. (ooo vs ooo = balanced; xo not weighed --> weigh the xo not weighed --> heavier ball is x)

If the 3 vs. 3 is imbalanced, then you know that the bomb is one of the 3 on the heavier side of the scale (ooo vs. oox = imbalanced). From the 3 balls on the heavier side of the scale (oox) weigh 1 random ball against another random ball. If the two sides balance (o vs o; x not weighed), then you know that the unweighed ball is the bomb. If the two sides do not balance (o vs. x) then you know that the ball on the heavier side of the scale is the bomb.

14. Aug 20, 2009

### Jimmy Snyder

And see to it that the scales you use are delicate. The weight of the explosives is small when compared to the weight of a ping pong ball. The terrorist was arrested when he purchased a package of band-aids to cover a small burn on one of his fingers from a test explosion gone out of control. Police were on the lookout and had alerted all of the grocery stores, pharmacies, convenience stores and public restroom vending machines.

15. Aug 27, 2009

### maverick_starstrider

Lol where did this person get this question? "Freedom Highschool"?

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