Can You Identify the Odd Ball in Three Attempts?

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The discussion centers on solving a classic puzzle involving eight balls, one of which is either heavier or lighter. Participants detail a systematic approach using a two-sided weight scale to identify the odd ball within three weighings. The method involves weighing pairs of balls and deducing the odd ball's weight based on the outcomes of the weighings. Additionally, the conversation touches on variations of the puzzle, including a cannonball problem that raises questions about projectile motion and air resistance.

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  • Familiarity with weighing scales and their use in problem-solving
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Recently, my friend tested me on a puzzle. The puzzle goes like this, if there is 8 balls (1 of them which weighs either lesser or heavier), how are you going to find out which is the different ball using a two sided weight scale with 3 tries?

This is a pretty easy puzzle which everyone would solve given enough time. After giving it some thought, I realized I can not only find which is the different ball with 3 tries but also tell whether it is heavier or lighter then the rest.

How do I do it?
 
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icyboy771z said:
Recently, my friend tested me on a puzzle. The puzzle goes like this, if there is 8 balls (1 of them which weighs either lesser or heavier), how are you going to find out which is the different ball using a two sided weight scale with 3 tries?

This is a pretty easy puzzle which everyone would solve given enough time. After giving it some thought, I realized I can not only find which is the different ball with 3 tries but also tell whether it is heavier or lighter then the rest.

How do I do it?

This is actually a very common puzzle. A bit of Googling will reveal the answer as well as many variations on it.
 
It commonly done with 12 balls.

number the balls

weigh 1,2 against 3,4
1. if they balance, the 5,6,7,8 has the odd one - weigh 123 vs 567
- if 123 vs 567 balance then 8 is the odd one (weigh against any other for heavy/light.
- if 567 is heavy, then they contain the heavy ball, weigh 5 vs 6: if 5 vs 6 balance, then 7 is heavy otherwise the heavier of 5,6 is the odd ball.
- if 567 is light then they contain the light ball - as above.

2. 1,2 vs 3,4 not balanced then 5,6,7,8 are all the same - weigh 1,2,3 vs 5,6,7
- if balanced then 4 is the odd one
- if 123 > 567 then the odd one is heavy, weigh 1 vs 2
- if 123 < 567 then the odd one is light, weigh 1 vs 2

Try this one:
http://trickofmind.com/2008/11/cannon-ball-conundrum.html

A cannon ball is shot out of a cannon and travel 4 miles before it hits the ground, at the same time a cannon ball is dropped out of the sky 4 miles off the ground

The question is Which Cannon ball will hit the ground first, Explain


I've seen it in different versions, here's a discussion:
http://simonbjournal.blogspot.com/2011/06/cannon-balls.html
 
Last edited:
Simon Bridge said:
It commonly done with 12 balls.

number the balls

weigh 1,2 against 3,4
1. if they balance, the 5,6,7,8 has the odd one - weigh 123 vs 567
- if 123 vs 567 balance then 8 is the odd one (weigh against any other for heavy/light.
- if 567 is heavy, then they contain the heavy ball, weigh 5 vs 6: if 5 vs 6 balance, then 7 is heavy otherwise the heavier of 5,6 is the odd ball.
- if 567 is light then they contain the light ball - as above.

2. 1,2 vs 3,4 not balanced then 5,6,7,8 are all the same - weigh 1,2,3 vs 5,6,7
- if balanced then 4 is the odd one
- if 123 > 567 then the odd one is heavy, weigh 1 vs 2
- if 123 < 567 then the odd one is light, weigh 1 vs 2

Try this one:
http://trickofmind.com/2008/11/cannon-ball-conundrum.html

A cannon ball is shot out of a cannon and travel 4 miles before it hits the ground, at the same time a cannon ball is dropped out of the sky 4 miles off the ground

The question is Which Cannon ball will hit the ground first, Explain


I've seen it in different versions, here's a discussion:
http://simonbjournal.blogspot.com/2011/06/cannon-balls.html

with regards to the cannonball problem, the "travel 4 miles" is problemmatic. is this the arc-length of the projectory, vertical travel, or horizontal travel?

and of course, there's variations in air density, coefficients of drag, the fact that 4 miles is enough away from the surface of the Earth to affect the acceleration, it's...ugly.

if it is one of the famous acme frictionless cannonballs, and the various vagaries of drag, air density, non-constant acceleration, etc. are ignored, AND we only mean the vertical travel of the cannonball, then they hit at the same time.
 
Yeah - that's why I linked to my earlier observations along the same lines.

Some versions have the height at 10 miles. The large heights lead me to suspect that the cannon firing may be intended as a red herring - the lynchpin may be to realize that here is a situation where it does not matter how the cannon is fired. The dropped ball starts so far up it always hits last?

Retellings garbled the numbers.

Have a google about - you'll see it keeps cropping up as a brain-teaser.
 

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