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

Actually, you can do up to 40 balls in four weighings, not just 39.

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 :-) |

Guest Message by DevFuse

Started by rookie1ja, Mar 30 2007 05:53 PM

35 replies to this topic

Posted 24 March 2010 - 09:01 PM

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

Actually, you can do up to 40 balls in four weighings, not just 39.

Posted 30 September 2010 - 06:40 AM

Didn't see it in the posts, i do miss things, but an alternative way to the solution for the situation where the 1234 = 5678 on the first weight, if you weighed 1,2 against 9,10 if equal then either 11 or 12 then weigh 11 against 1 if equal then 12 is odd and if unequal then 11 is odd but if 1,2 against 9,10 is unequal weigh 1 against 9 >> if equal 10 is odd if not 9 is odd, mark

Posted 13 February 2011 - 12:44 PM

The least number of times is 4, according to me.

First separate the balls into 3 groups of 4 balls each.

1st weighing - group 1 against group 2

2nd weighing - group 1 against group 3

By doing this, you know which two are equal. Consider the group which is heavier or lighter...

Divide this group into two, X and Y. Also take two balls which u are certain are of normal weight.

3rd weighing - X against 2 normal balls

If unequal, oddball is in X.

If equal, oddball is in Y.

Let's for now assume that the oddball is in Y.

4th weighing - one of the balls in Y against a normal ball.

If unequal, that's your ball... if equal, the other ball in Y is the oddball

First separate the balls into 3 groups of 4 balls each.

1st weighing - group 1 against group 2

2nd weighing - group 1 against group 3

By doing this, you know which two are equal. Consider the group which is heavier or lighter...

Divide this group into two, X and Y. Also take two balls which u are certain are of normal weight.

3rd weighing - X against 2 normal balls

If unequal, oddball is in X.

If equal, oddball is in Y.

Let's for now assume that the oddball is in Y.

4th weighing - one of the balls in Y against a normal ball.

If unequal, that's your ball... if equal, the other ball in Y is the oddball

Posted 16 May 2011 - 02:11 AM

tambay, i think you were finding out how to tell if that specific one is lighter or heavier, and even then, you'd have to have more steps to do that. the next step from case 1/2 is to take the other ball and see if it balances with one of the balls from case 1.if it does, then you have to try the other set of 3 to find out if that specific one is lighter or heaver by repeating the step that lead to case1/2.I have a shorter solution.

Put six balls on each side of the scale. Whichever is lighter, the lighter ball is in that group of six.

From these balls, put three balls on each side of the scale.

Whichever is lighter, the lighter ball is in that group of three.

From the three balls, put a ball on each side of the scale.

Case1: If the scales balance, the third ball is the lighter ball.

Case2: If one is lighter, then that is the ball...

you would do the same for case 2, and if the two balls balanced (leaving one on the scale and replacing the other), then you know that the one you replaced is heavier

my "case 3" would tell you if its lighter (take the step that lead to case 1/2 and replace one of the 2 balls, and if they balance, then the lighter one is the one you replaced, and to clarrify, NOT the one you replaced it with)

Posted 07 June 2011 - 01:05 PM

Knowing what the site admin said, you could just rephrase this asI have a shorter solution.

Put six balls on each side of the scale. Whichever is lighter, the lighter ball is in that group of six.

From these balls, put three balls on each side of the scale.

Whichever is lighter, the lighter ball is in that group of three.

From the three balls, put a ball on each side of the scale.

Case1: If the scales balance, the third ball is the lighter ball.

Case2: If one is lighter, then that is the ball...

I have a shorter solution.

Put six balls on each side of the scale. Whichever is lighter or heavier, the lighter/heavier ball is in that group of six.

From these balls, put three balls on each side of the scale.

Whichever is lighter/heavier, the lighter/heavier ball is in that group of three.

From the three balls, put a ball on each side of the scale.

Case1: If the scales balance, the third ball is the lighter ball.

Case2: If one is lighter, then that is the ball...

Posted 08 June 2011 - 03:52 AM

I have a shorter solution.

Put six balls on each side of the scale. Whichever is lighter, the lighter ball is in that group of six.

From these balls, put three balls on each side of the scale.

Whichever is lighter, the lighter ball is in that group of three.

From the three balls, put a ball on each side of the scale.

Case1: If the scales balance, the third ball is the lighter ball.

Case2: If one is lighter, then that is the ball...

Spoiler for Remember

0 members, 0 guests, 0 anonymous users