This is a very old Logic Problem from one of my Professors when I was in school. I've modified it slightly to be of interest to those of us who Reload.

...

You have a Balance Beam Scale which has a 6" diameter Pan on each side.

And you have 12 - 0.308" 165gr Bullets which look as if they are all exactly the same. Eleven of them have the EXACT same weight and one is either slightly heavier or lighter.

The difference in weight can not be felt. You do not have access to any other objects of a known weight.

What is the minimum number of weighings it takes to know for absolute certainty which Bullet is different and if it is heavier or lighter?

...

It took me 30 minutes just to remember the Problem. If you do figure it out, I'll encourage you to hold onto the way to do it until everyone else that expresses an interest in "thinking through it" has responded, but do respond with the number of weighings it took you.

Best of luck to you folks.

Depending on luck it could be two or it could be three.
Jim

I keep comming up with a min of 1 (luck) and a Max of 5.
How'd I do?

Answer for certain in three (3) is what I come up with when using a balance beam, 6/6, 3/3, 1/1.



After a bit more thought and a re-read (one it lighter OR heavier) I believe four (4) will have it for certain. 6/6, 3/3, 3/3, 1/1

min of 2

1

THREE...BCB

Min with luck has to be 2.. Remember the question is weather its lighter OR heavier!!! So if by luck you got 2 of a differnet weight on the first weighing you still dont know which is the oddball! All that will tell you is one is heavier then the other!!! So are the other ones the weight of light one or the heavy one???

First of all, you would need to determine whether the oddball bullet is light or heavy, then you need to find it. Luck is the only way this could be achieved with one weighing.

From a diagnostic approach I come up with 3 minimum.

Hot Core,

Being anally retentive, I would use all 12! Then there is NO statistical possibility of error!

ASS_CLOWN

Oops-I made the mistake of assuming that you were told whether the oddball was lighter or heavier.Not knowing which would require another use of the scale making the answer four.

i get 5 to be sure.
woofer

Four for certain

A "divide and conquer" approach, if you will, makes 4 the answer.

Has to be three for both answers.

Minimum of 2

I came up with four. There can be no "Luck" for this problem. Also, If you have a drawer full of half white socks and half black socks and the room is dark, how many socks must you take out to be Sure to have one pair?
If you have a glass that holds 12 oz. and there is 6 oz. of beer in the glass, is the glass 1/2 full or 1/2 empty? None of the above, You have a container that is twice as big as the amount of material you have. Sorry, got carried away!!!.

The minimum is 2.

The answer to Steve4102's question is three.

I'll bet your house against mine I can do it in 3. No less, no more.

6/6,3/3,2/2,1/1

In that case I get 3, in the unknown case (where the defective bullet may be heavier or lighter I get 4 - changed my mind). Here's my logic:

Weigh 4/4 (leave one out). One side will be lighter (if not, then the one left out is the light one). Take the 4 on the light side and weigh 2 and 2. Again take the two from the lighter side. Now weigh these 1/1 to get which is lighter. Three weighings.

If we do not know if the bullet is light or heavy, then we need to make one additional weigh. Here is my logic:

Weigh 4/4, one side is heavier, but we do not know if it heavier because of a defective light bullet on one side or because of a defective heavy bullet on the other. Thus, we weigh both subsets of 2/2, one of which will be an equality, which then determines the nature of the defect. Then weigh 1/1 and you are done. Four weighings. A similar logic works for 12, with one extra step, but still only 4 weighings.

I could be wrong - these logic questions always have a loophole somewhere. Fun though...

Yes, that is right. And then one additional weighing if the defect is unknown, for a total of three. Uh-huh - I guess problems with an even # of bullets are more my forte

I think it is 3 for 12 bullets, and two for 9.

Hotcore said 12 bullets, not nine...

you would have to weigh 6/6, that is weighing number 1

one side of the scale will drop, the other will rise

then you remove 3 from each side

if the scale balances, throw those bullets off and put the ones back on that you removed. if the scale does not balance set the six that you removed aside. that's weighing number 2 either way

there will now be three bullets in each pan. either a heavy one is in the left pan or a light one is in the right pan

now you would pull two bullets from each pan. if the scale stays the same that means the four bullets you are holding weigh the same so you set them aside in the 'normal' pile

so now you have to substitute one of the normal bullets you have set aside for one of the bullets on the scale. if the bullet you switch causes the scale to balance then you know you have removed the defective bullet and depending on which way the scale moved you know if that bullet is light or heavy

that's weighing number 3 and you could be finished now.

but what if when you remove those two bullets from each pan the scale balances?

then you would know that of the four bullets you are holding that one of them is heavy or light. throw the two bullets still on the scale on the 'normal' pile. you haven't used the third weighing yet.

you will have to take the four bullets left (one of these is the odd one) and put two on each side of the scale. the scale will tilt to one side. you would then remove one bullet from each side of the scale and either the scale will balance or it won't. this is weighing number 3 but you are not finished yet

what you want now is for the scale NOT to balance so if the scale balances when you remove one bullet from each side in the above step you just remove those two normal bullets from the pans and put the imbalanced bullets back (you could call this weighing number 4 but all you will really be doing is putting back the imbalanced bullets).

one of the two bullets remaining on the scale is either heavy or light. all you have to do is take a normal bullet and replace one of the bullets on the scale. this I would call weighing number 4 but as i said above you might take issue with that and say it is 5.

anyway the 'normal' bullet you replace one of the two remaining bullets with will cause one of two things to happen.

either the scale balances and that means the bullet you took away is light or heavy depending on whether the scale moved up or down to level.

or the scale stays the same and that means that the remaining bullet is either light or heavy depending on whether it is up or down

i say four weighings, five if you count the replacement of the two bullets in that earlier step

dhs

Whether the number of bullets is 9 or 12,I would divide the number of bullets into three even groups and start from there.

the answer is four weighings to both determine Which bullet is DIFFERENT and wheather it is heavy or lite.

I sure want to "blow my horn" but respect your request to wait for the others "smoke" their brains, first. From the 30 posts currently there is some scourched gray matter.

edit...



where's my manners?



sorry, hotcore you asked us not to give out the solution



dhs

TWO





Hint: set six aside and weigh 3 vs 3 first



Hint: These are PANS and sensitive to the POSITION of the bullets on them.



Hint: Adding 165 grs to the underweight side increases the total weight by a greater proportion than adding 165 grs to the overweight side.

If I dropped all 12 bullets onto a Remington 700 safety, would I get an accidental discharge?

Hot Core,

No, I am saying I hate statistics so I would use all 12 bullets, and 11 individual weighings (MAX). I guarantee it would irrefuteable proof.

ASS_CLOWN

Hey Squeezer, It appears that you are only using " 3 " of the " 12 " bullets.

Have I misread that? A=1, B=2, C=3

No " LUCK " needed or involved!