• We’re currently investigating an issue related to the forum theme and styling that is impacting page layout and visual formatting. The problem has been identified, and we are actively working on a resolution. There is no impact to user data or functionality, this is strictly a front-end display issue. We’ll post an update once the fix has been deployed. Thanks for your patience while we get this sorted.

Riddle: You have 10 bags...

Page 3 - Seeking answers? Join the AnandTech community: where nearly half-a-million members share solutions and discuss the latest tech.
Originally posted by: Gobadgrs
Originally posted by: silverpig
Here's a tough one that took me a few hours to work through. Please don't post spoilers for the people who want to solve it

You have 12 golf balls, 11 of the same weight and one of a deviating weight. You are allowed to use a balance scale 3 times. BUT, you do not know whether the deviating golf ball is heavier or lighter than the rest. Using the scale 3 times, you have to point out the deviating ball AND tell whether its heavier or lighter than the rest.

Thats so frickin easy.

Go ahead and post your solution. It's actually a fairly difficult problem if you haven't seen it before.
 
And actually I believe there are two independent unique solutions to my riddle. I found one on my own in a few hours. It's definitely a tough one.
 
what are you doing, practicing for consulting interviews?

I disticntly remember seeing this in the vault guide, something about a king and ppl bringing bags of gold
 
Originally posted by: ndee
Originally posted by: Triumph
Originally posted by: ndee
Solution as some have found:

Weigh one coin from the first bag, two coins from the second bag, three coins from the third bag, etc.
100 times the difference from 55 grams is the bag that weighs 99 grams.

copyright by chuckywang

um, that's more than one weighing.

ah, it was already late. Ok, here's the real solution:

Take 1 coin from the first bag, 2 coins from the second, 3 coins from the third, etc. Then, put them all together on the scale -> with that difference to 55, you find out which bag weighs 99g.

or yeah, you weigh them all at once, I think that's what chuckywang meant.


Clarify.

From what you say you're left with one bag of 99 coins through to the last bag with 90 coins in it. Plus you have an eleventh bag with 55 coins in it.

Then what? you put them all on the scale? It's still just going to come to the same weight.

Explain better.
 
Didn't read it to ifnd the answer but you number the bags 1 - 10 and take that many coins from each bag. 1 from 1, 2 from 2, etc. You put all those coins on the scale and see what it weighs. That's a total of 55 coins. Take the weight on the scale and however many tenths it is below 55 is the bag that has the lighter coins...
 
Originally posted by: silverpig
Bump. Anyone solve mine yet? 🙂


My method only narrowed it down to two, and didn't tell if the oddball was heavier or lighter. Then I cheated and looked up an answer. 😉
 
Originally posted by: FrankyJunior
Silverpig: yours is easy if you know the ball is either havier or lighter but I'm not sure if you don't know....

It IS solvable... and that's the hard part 🙂
 
i googled the solution to the one from silverpig posted, and you guys saying "it's easy" are full of dogshit.
 
Originally posted by: Kev
i googled the solution to the one from silverpig posted, and you guys saying "it's easy" are full of dogshit.

yeah, i thought it was easy at first until i saw the "heavier or lighter" caveat. i don't know the answer.
 
The solution would have been relatively simple for me, yet given the information as stated in the original post, it's unsolvable.
Nowhere does it say that each coin in a bag has an identical mass as each other coin in that same bag.

 
Originally posted by: loic2003
Originally posted by: ndee
Originally posted by: Triumph
Originally posted by: ndee
Solution as some have found:

Weigh one coin from the first bag, two coins from the second bag, three coins from the third bag, etc.
100 times the difference from 55 grams is the bag that weighs 99 grams.

copyright by chuckywang

um, that's more than one weighing.

ah, it was already late. Ok, here's the real solution:

Take 1 coin from the first bag, 2 coins from the second, 3 coins from the third, etc. Then, put them all together on the scale -> with that difference to 55, you find out which bag weighs 99g.

or yeah, you weigh them all at once, I think that's what chuckywang meant.


Clarify.

From what you say you're left with one bag of 99 coins through to the last bag with 90 coins in it. Plus you have an eleventh bag with 55 coins in it.

Then what? you put them all on the scale? It's still just going to come to the same weight.

Explain better.


ok, for this example, lets say the 4th bag is the bag with the lighter coins.

#1: 1g x1 = 1g
#2: 1g x2 = 2g
#3: 1g x3 = 3g
#4: .99g x4 = 3.96g
#5: 1g x5 = 5g
#6: 1g x6 = 6g
#7: 1g x7 = 7g
#8: 1g x8 = 8g
#9: 1g x9 = 9g
#10: 1g x10 = 10g

total weight = 54.96g

55g-54.96g=.04 x 100 = 4 therefore bag #4 has the lighter coins.

you can trade in other bags and whatever the difference is, multiply by 100 and you will get the bag with the lighter coins.


 
Originally posted by: vood0g
Originally posted by: loic2003
Originally posted by: ndee
Originally posted by: Triumph
Originally posted by: ndee
Solution as some have found:

Weigh one coin from the first bag, two coins from the second bag, three coins from the third bag, etc.
100 times the difference from 55 grams is the bag that weighs 99 grams.

copyright by chuckywang

um, that's more than one weighing.

Crazy. I don't know how anyone would think of that.

ah, it was already late. Ok, here's the real solution:

Take 1 coin from the first bag, 2 coins from the second, 3 coins from the third, etc. Then, put them all together on the scale -> with that difference to 55, you find out which bag weighs 99g.

or yeah, you weigh them all at once, I think that's what chuckywang meant.


Clarify.

From what you say you're left with one bag of 99 coins through to the last bag with 90 coins in it. Plus you have an eleventh bag with 55 coins in it.

Then what? you put them all on the scale? It's still just going to come to the same weight.

Explain better.


ok, for this example, lets say the 4th bag is the bag with the lighter coins.

#1: 1g x1 = 1g
#2: 1g x2 = 2g
#3: 1g x3 = 3g
#4: .99g x4 = 3.96g
#5: 1g x5 = 5g
#6: 1g x6 = 6g
#7: 1g x7 = 7g
#8: 1g x8 = 8g
#9: 1g x9 = 9g
#10: 1g x10 = 10g

total weight = 54.96g

55g-54.96g=.04 x 100 = 4 therefore bag #4 has the lighter coins.

you can trade in other bags and whatever the difference is, multiply by 100 and you will get the bag with the lighter coins.



Crazy...
 
Originally posted by: DrPizza
The solution would have been relatively simple for me, yet given the information as stated in the original post, it's unsolvable.
Nowhere does it say that each coin in a bag has an identical mass as each other coin in that same bag.
Exactly, I hate poorly written "logic" questions. What if that lighter bag had 98 coins that each weighed 1 gm and two coins that each weighed 0.5 gm. That is a problem which is unsolvable with just one weighing. We should NOT assume that the coins all weigh the same unless told in the first post.
 
Originally posted by: silverpig
Here's a tough one that took me a few hours to work through. Please don't post spoilers for the people who want to solve it

You have 12 golf balls, 11 of the same weight and one of a deviating weight. You are allowed to use a balance scale 3 times. BUT, you do not know whether the deviating golf ball is heavier or lighter than the rest. Using the scale 3 times, you have to point out the deviating ball AND tell whether its heavier or lighter than the rest.
I had one class where we were given ~7 variations of this same problem. If we solved them all, we got to skip one exam. I had an A at the end of the regular class and didn't want to take the final. So I did those problems and got to skip the final. A couple hours with these problems was far nicer than many hours studying.
 
Originally posted by: chuckywang
Originally posted by: Kyteland
Originally posted by: silverpig
Bump. Anyone solve mine yet? 🙂
Yes. Want to know the answer? 😉
I almost got it. But it required that one of the first two weighings must have been off balance.
Alright, I'm going to have to post my solution now. It took almost as much time to type up as to solve. 😱

Don't read the below quote if you still want to figure this out yourself.
Damn, this was a pain to write up.

Solution and SPOILER:

Label the balls A-L. Each weighing one of three things happens. 1) the left side of the scale moves (U) up, 2) the left side of the scale moves (D) down, 3) the scales are (=) equal.

-Weigh ABCD vs EFGH

-If U: the unknown is in ABCDEFGH
-Weigh ABCEF vs DIJKL
--If U: the unknown is in ABC and lighter
--Weigh A vs B
---If U: the unknown is A and lighter
---If D: the unknown is B and lighter
---If =: the unknown is C and lighter
--If D: the unknown is in DEF
--Weigh E vs F
---If U: the unknown is F and heavier
---If D: the unknown is E and heavier
---If =: the unknown is D and lighter
--If =: the unknown is in GH and heavier
--Weigh G vs H
---If U: the unknown is H and heavier
---If D: the unknown is G and heavier

-If D:
This is the same case as the U case only reversed.

-If =: the unknown is in IJKL
-Weigh IJK vs ABC
--If U: the unknown is in IJK and lighter
--Weigh I vs J
---If U: the unknown is I and lighter
---If D: the unknown is J and lighter
---If =: the unknown is K and lighter
--If D: the unknown is in IJK and heavier
--Weigh I vs J
---If U: the unknown is J and heavier
---If D: the unknown is I and heavier
---If =: the unknown is K and heavier
--If =: the unknown is L
--Weigh L vs A
---If U: the unknown is L and lighter
---If D: the unknown is L and heavier

Edit: After consulting Google that seems like a different answer than other people got. Oh well, it still works.
 
That's pretty good Kyteland. Here's the solution I came up with spoiler warning

Divide them up into 3 groups of 4. Call them Group A, B, and C.

Weigh Group A vs Group B.



Case 1: Group A = Group B.

In this case, you know the deviant ball is in group C. Weigh c1, c2 and c3 against a1, a2, and a3.



case 1.1: They are the same.

You know the deviant ball is c4. Weigh it against a1 to find out if it is light or heavy.

case 1.2: c1, c2 and c3 are heavier.

Weigh c1 vs c2. If they are the same, you know the ball is c3 and it is heavier. If they differ, you know that the heavier ball is the one.

case 1.3: c1, c2, and c3 are lighter.

Same as above. Weigh c1 vs c2. If they are the same, you know c3 is the light ball. If they differ, the ball that is lighter is the one you're after.




Case 2: A is heavier than B.

Take a1, a2, and b1, and weigh them against a3, b2, and c1.



case 2.1: a1, a2, and b1 are heavier.

Since you know the heavy ball must be in A or the light ball is in B, you can tell that the ball you want is either a1, a2, or b2. Weigh a1 vs a2.


case 2.1.1: They differ.

You know that the heavier ball is your ball.

case 2.1.2: They are the same.

You know that b2 is your ball and it is lighter.



Case 2.2: a1, a2, and b1 are lighter.

You know either b1 is your ball and it is lighter, or a3 is your ball and it is heavier. Weigh a3 vs c1.


case 2.2.1: a3 is heavier than c1.

You know a3 is your ball and it is heavy.

case 2.2.2: a3 is the same as c1.

You know b1 is your ball and it is lighter.



Case 2.3: a1, a2, and b1 are the same weight as a3, b2, and c1.

Basically, at this point, you have a4, b3, and b4 left. From your first measurement, you also know that a4 cannot be the lighter ball, nor can b3 nor b4 be the heavier ball. Take a4 and b3, and weigh them against c1 and c2.


case 2.3.1: a4 and b3 are heavier.

Since you know that b3 cannot be the heavier ball, a4 must be the ball you want, and it is heavy.

case 2.3.2: a4 and b3 are lighter.

Since you know that a4 cannot be the lighter ball, b3 must be the ball you want and it is lighter.

case 2.3.3: a4 and b3 are the same as c1 and c2.

You know that b4 is the deviant ball, and from your very first measurement, you know that it must be light.


Case 3: B is heavier than A:

Just repeat case 2 over, but switch all the balls labelled a for b.

DONE
 
Originally posted by: DrPizza
The solution would have been relatively simple for me, yet given the information as stated in the original post, it's unsolvable.
Nowhere does it say that each coin in a bag has an identical mass as each other coin in that same bag.

 
Back
Top