# Can You Solve This Riddle?

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References 🗂

This riddle first appeared in Grossman, Howard (1945). Scripta Mathematica XI

TedED have a similar riddle that’s beautifully animated: https://youtu.be/tE2dZLDJSjA

source

Fahad Hashmi is one of the known Software Engineer and blogger likes to blog about design resources. He is passionate about collecting the awe-inspiring design tools, to help designers.He blogs only for Designers & Photographers.

### 44 thoughts on “Can You Solve This Riddle?”

• December 4, 2017 at 2:11 am

The solution? Ask Ted-Ed. They have it.

• December 4, 2017 at 2:11 am

Damn it i cant find my marbles

• December 4, 2017 at 2:11 am

How to balance 10 nails on the head of one nail.

• December 4, 2017 at 2:11 am

Damn I alwas put 5 vs 5 from the frist try to know where it is DAMN ME… alwas do that it just sond so logical after I know what I am looking for but whit 4 you can know what you are looking for too because the one outside the scale are more or less on the scale too (if you see the scale is balance whit the 4 vs 4 you know it in the 4 outside of the scale It give you more info.)
I just don't know I alwas pref going strait to 50% 50% whit that I can split them in 2 again etc.

• December 4, 2017 at 2:11 am

Considering it can be either heavier or lighter, a different approach:

3 sets of marbles: ABCD – PQRS – WXYZ

if <ABCD> vs <PQRS> is balanced
|………if <WX> vs <AB> is balanced
|………|………if <Y> vs <A> is balanced
|………|………|………The different is Z
|………|………else
|………|………|………The different is Y
|………|………end if
|………else
|………|………if <W> vs <A> is balanced
|………|………|………The different is X
|………|………else
|………|………|………The different is W
|………|………end if
|………end if
else
|………if <ABPQ> vs <CRYZ> is balanced
|………|………if <D> vs <Z> is balanced
|………|………|………The different is S
|………|………else
|………|………|………The different is D
|………|………end if
|………else
|………|………if the scale tilts to the same side as before
|………|………|………if <A> vs <B> is balanced
|………|………|………|………The different is R
|………|………|………else
|………|………|………|………if the scale tilts to the same side as before
|………|………|………|………|………The different is A
|………|………|………|………else
|………|………|………|………|………The different is B
|………|………|………|………end if
|………|………|………end if
|………|………else
|………|………|………if <P> vs <Q> is balanced
|………|………|………|………The different is C
|………|………|………else
|………|………|………|………if the scale tilts to the same side as before
|………|………|………|………|………The different is P
|………|………|………|………else
|………|………|………|………|………The different is Q
|………|………|………|………end if
|………|………|………end if
|………|………end if
|………end if
end if

• December 4, 2017 at 2:11 am

Enough to cause me to lose my marbles!

• December 4, 2017 at 2:11 am

Somebody please show this to Captain Holt from Brooklyn Nine Nine please.

• December 4, 2017 at 2:11 am

This is the same as the coin riddle from Teded.

• December 4, 2017 at 2:11 am

I'm glad you like making videos.
I like watching the videos you make. 🙂

• December 4, 2017 at 2:11 am

I got the part about grouping the marbles into three parts, but I had the thought that if you knew that 11 out of 12 of the marbles were equal in mass that you could divide the heavier or lighter group into two groups with two balls and then deduct that way because you know that 3 out of the 4 balls are going to be equal in mass to all the other balls. I then realized your method would always take three times whereas mine might take four or five, and I always solve these riddles almost correctly.

• December 4, 2017 at 2:11 am

I came up with a different solution. To explain, I'll assign each marble a letter (A, B, C, D, E, F, G, H, I, J, K, L), and number to show if is the same as the rest (0), might be lighter than the rest (1), heavier than the rest (2), or unknown (3).

We start with a set of all twelve unknown marbles: [A3, B3, C3, D3, E3, F3, G3, H3, I3, J3, K3, L3]

I weigh any two groups of four: [A4, B4, C4, D4] vs [E4, F4, G4, H4]
From here one of three things will happen. The scale will be balanced, the scale will tilt left, or the scale will tilt right.

If the scale is balanced, then I do the same as you did in the video.

If the scale is tilted then I know that the other four are the same.
From here I will set two of the heavy side marbles to the side, and weigh two light and one heavy marble against the other two light, and the other heavy marble: [A1, B1, E3] vs [C1, D1, F3]

If this is balanced, then I know that they're all the the same, and can easily find the heavy marble from the two I set aside by simply weighing them against each other: [G3] vs [H3], and the heavier one is the odd one out.

If the second weigh is tilted, then I know that the odd one out is among either the heavy marble on the lower side, or the light marbles on the higher side. Thus, I can simply weigh the two remaining light marbles against each other [A1] vs [B1]
If this is balanced, then the remaining heavy marble is the odd one out. Otherwise, it is the lighter of the two.

This works every time, with one of 24 possible results.

• December 4, 2017 at 2:11 am

My favorite teaser is the grandfather paradox. So if you have the time you can expound it on this video and provide your own solution to the grandfather paradox.

• December 4, 2017 at 2:11 am

Still won't work because between the last two marbles, you still won't know which is the odd one out… Am I wrong?

• December 4, 2017 at 2:11 am

My favourite puzzle. My maths teacher posed this to the class when I was 12. He offered a crate of cola to anyone who could do it. I won the crate.

• December 4, 2017 at 2:11 am

I didn't solve it completely, I solved the first part (in another way)
but i couldn't find such a way to solve if the balance is unbalanced. Thank you for this amazing video 😉

• December 4, 2017 at 2:11 am

My method starts the same but is different.
First divide it in 3 groups of 4 marbles (group A, group B and group C)
1) Weigh group A against group B, if they are the same we know the different one is in group C so
2) Weigh 3 of group A against 3 of group C, if they are the same we know the one of group C that was left out is the different one. And if we want to know if it's heavier or lighter just
3) Weigh it against any one of the others to see if it’s heavier o lighter.
If in the second weigh (3 A against 3 C) they weigh different, then we know the different marble is in those 3 from group C (and depending on if that group C was lighter or heavier we will know if the one that’s different is lighter o heavier) so
3) Weigh one of that group against another one of that same group, if they weigh the same the one that’s left out is the different one, if they weigh different we will know which one is the different one because on the second weigh we found out if the different one was heavier or lighter.
If in the first weigh group A and group B weigh different, then let’s take the heavier group (let’s assume for this example it’s group A) and
2) Weigh 3 from group A against 3 from group C, if they weigh the same then we know the different one is either the one left out from group A or in group B so
3) Weigh the one left out from group A with 2 from group C against 3 from group B, if they weigh the same we know the one left out from group B is the different one, and it’s lighter. If the side with the one from group A and 2 from group C weighs more then the different one is that one from group A and it’s heavier. The side with 3 from group B can’t weigh more.
If the result from weighing 3 from group A against 3 from group C is that the side with group A weighs more (the side with group C can’t weigh more) then we know that the different one is among those 3 from group A, so
3) Weigh one of that group A against another one of that same group, if they weigh the same the one that’s left out is the different one, if they weigh different we will know that the one that’s heavier is the different one.
(If in the first weigh group A and group B weigh different and the heavier group is group B instead of group A the steps that follow are the same just inverting A with B)

• December 4, 2017 at 2:11 am

Weigh 3 against 3. If it is one of those piles weigh 1 against 1. If it is not either of the first set of 3s, weigh the other group's 2 against 2.

I win.

• December 4, 2017 at 2:11 am

Without completing the video I think I've worked it out. Split the marbles into groups of 4. Weigh one against another. The heavy marble will be in the group that weighs more. If they weigh the same it's in the unused group. From here it's as easy as weighing 2 against 2 and then 1 against 1.

• December 4, 2017 at 2:11 am

you don't know which is the odds one is the odds one out you just know there is a difference on weighting one against another.

• December 4, 2017 at 2:11 am

Take a drink whenever she says marble.

• December 4, 2017 at 2:11 am

i did it with a different method.
1) 4 Vs 4 ( further steps same as yours if balanced)
( if unbalanced)
2) HHN Vs HHL

• December 4, 2017 at 2:11 am

There is also this one: A cowboy rode into town on Friday, he stayed for 3 days and left on friday. How is this possible?

He has a horse named Friday.

• December 4, 2017 at 2:11 am

Ok this is a freebie for everyone who sees this comment! Though this brain teaser is best if said out loud:

One knight, a king and queen enter a castle. The castle was completely empty, no one else entered or left the castle. Then the king and queen left the castle. But there was actually a third person, who was it?

Answer: The knight. The explaination for this is because the brain teaser is not talking about night time, it is talking about a knight as in the type of soldier.

Try this on other people, and see if they get it! 🙂

• December 4, 2017 at 2:11 am

well, here is one that I think is good and took sometime to solve. you have ten packs of chewing gum ( the original one was cigarettes) each pack contains 10 gums and each gum weighs 10, except one of the packs each gum weighs 9 gm. all what we have is a sensitive scale which shows the weight by grams. how can you tell the flawed pack with only one use of the scale?

• December 4, 2017 at 2:11 am

I thought there is 11 marbles total, had a great time

• December 4, 2017 at 2:11 am

10 things l know about you:
2. You're a human.
3. You've just pressed read more.
4. You can't say the letter ' p ' without separating your lips.
6. You just attempted to do it.
7. You 're laughing at yourself.
8. You have a smile on your face and you just skipped number 5.
9. You just checked to see if there is number 5.
10. You're laughing at this because everyone does it.
11. You'll probably copy and paste this and make it your thingy, too.
12. You are wondering why there is number 11 and 12.

• December 4, 2017 at 2:11 am

I'm no audio expert, but I believe you are clipping your mic. I believe this is why people put those mesh in front of the mics.
Try putting a cloth in front of the mic to help.

• December 4, 2017 at 2:11 am

1+11, 3&9 and 6&6? So the groupings include when its appreciable?

• December 4, 2017 at 2:11 am

When you're handling the marbles to put them in and take them out of the scales, be alert for one which feels heavier or lighter in your hand than the rest. That's the one you're looking for.

• December 4, 2017 at 2:11 am

LOL, "Pause the video…" … two days later …. I unpaused
I am back to confirm my answer is correct ….. annnnnnnnnd …
Vanessa, you are right, there is more than one path to answer …
A MOST interesting and entertaining video … THANK YOU Vanessa!

• December 4, 2017 at 2:11 am

why can't you just place one marble on each side of the scale. If the scale is balanced you know that both marbles are normal; keep replacing one of the marbles til you get the odd weighing which will also determine whether the odd marble is heavier or lighter.
If the scale tips in the first weighing you know that there is a odd marble and normal, you can then replace both and prove that you have two normal marbles on a balanced scale, now you only have to replace one of the normal marbles with the one from the two weighed first time round, if the scale tips you have found your odd one, if it doesn't it's the only one left.
Simples.

• December 4, 2017 at 2:11 am

will I used the engineers​ logic the scale would be hard and not efficient so through away the scale get a long glass cup with car engine oil toss the 11 marbles at the same time in the cup one time for measuring two times to check 10 marbles should have the same velocity the lower or heavier mass marble would have a different velocity that's it fox problem solved 😉😉😉

• December 4, 2017 at 2:11 am

0:40 just take the two blue identical marbles and weigh them … all the other marbles don't look the same as a second marble

• December 4, 2017 at 2:11 am

I'm fond of the Monty Hall problem. An oldie but a goodie.

• December 4, 2017 at 2:11 am

• December 4, 2017 at 2:11 am

The Stop-Motion thing although looks cute, makes it really really hard to follow.
The marbles were just jumping around with no way of knowing how they're being moved.