## Welcome to BrainDen.com - Brain Teasers Forum

Welcome to BrainDen.com - Brain Teasers Forum. Like most online communities you must register to post in our community, but don't worry this is a simple free process. To be a part of BrainDen Forums you may create a new account or sign in if you already have an account. As a member you could start new topics, reply to others, subscribe to topics/forums to get automatic updates, get your own profile and make new friends. Of course, you can also enjoy our collection of amazing optical illusions and cool math games. If you like our site, you may support us by simply clicking Google "+1" or Facebook "Like" buttons at the top. If you have a website, we would appreciate a little link to BrainDen. Thanks and enjoy the Den :-) |

# Weighing IV.

### #21

Posted 22 February 2008 - 11:17 PM

### #22

Posted 25 February 2008 - 09:28 AM

The key to solving this (as "bonanova" shows) is realizing that you get more information out of the unbalanced scale than most people realize at first. Say you put to piles of four on either side of the scale. If it is uneven, you know that the odd ball is one of those eight (obvious). What wasn't as obvious (to me) at first was that you also know, that if the odd coin is later determined to be from the pile which is now on the low side of the scale, IT IS LIGHTER than the rest. Vice versa for the high side.

For whatever reason having to mix the "lighter than", "heavier than", and "known normal" balls was a no brainer once i realized this. Otherwise my solution was the same as Bonanova's noted above.

I really like the array solution though, brilliant Stephcorbin thanks!

### #23

Posted 12 March 2008 - 10:33 PM

### #24

Posted 25 April 2008 - 07:05 PM

### #25

Posted 20 May 2008 - 01:10 PM

### #26

Posted 04 June 2008 - 05:56 AM

There is a formula to solve this kind of problems( i got this formula from combinatorics, and found suitable for these kind of problems)

no. of uncertainties =x

no. of outcomes for a weigh =y

no. of weighings required =n

then y^n >x ( the minimum value of n satisfying the equation is the answer)

(this formula would be better for competitive exams)

for this problem;

no. of uncertainties = 2*12=24 ( each ball is heavier or lighter, we dont know)

no. of possible outcomes of each weigh=3 ( each weigh says whether one half is heavy,light or equal to other)

3^n>24 ( 3 is the min. value of n satisfying the equation)

therefore answer is 3.

**Edited by shobith, 04 June 2008 - 05:57 AM.**

### #27

Posted 07 September 2008 - 02:14 AM

start with 5 balls on either side and go from there. 3 tries will do it.To clarify:

The solution of weighing 6 and 6, 3 and 3, and then finding the oddball doesn't work. When you get one group to be heavier and one group to be lighter, you still don't know whether the oddball is heavier or lighter-- if it's heavier, you want to start weighing the heavy group, if it's lighter, you want to start weighing the lighter 6.

### #28

Posted 12 January 2010 - 03:11 PM

What would be the max number of balls you could have if you had 4 weighings? My first thought is more than 24 because if you weighed 6 v 6, worst case you would have 12 unknowns with 3 weighings left (same as this problem). Or you would have 6 possible heavys and 6 possible lights which is infomation you could use. So you could probably do 26 or something.

### #29

Posted 24 March 2010 - 07:36 PM

Could you do this same riddle with more balls, or is 12 the max?

Actually, it is possible to have up to 14 balls and determine which one is lighter/heavier in only three weighings. This makes the problem even more interesting; go ahead and try it

### #30

Posted 24 March 2010 - 08:07 PM

What would be the max number of balls you could have if you had 4 weighings? My first thought is more than 24...

The maximum number of balls for which you can solve the problem in 4 weighings is in fact 39.

If you have an additional ball that you know is "normal", you can do up to 41 balls. My previous reply assumed such an "extra" ball. Without the extra ball, the maximum number of balls for which you can solve the problem in three weighings is actually 13, not 14.

#### 0 user(s) are reading this topic

0 members, 0 guests, 0 anonymous users