## 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 :-)
Guest Message by DevFuse

# Weighing VII.

28 replies to this topic

### #1 rookie1ja

rookie1ja

Senior Member

• 1341 posts
• Gender:Male
• Location:Slovakia

Posted 30 March 2007 - 05:47 PM

Weighing VII. - Back to the Water and Weighing Puzzles
Given 27 table tennis balls, one is heavier than the others.
What is the minimum number of weighings (using a two-pan balance scale) needed to guarantee identifying the heavy one? Always. Of course, the other 26 balls weigh the same.

This old topic is locked since it was answered many times. You can check solution in the Spoiler below.
Pls visit New Puzzles section to see always fresh brain teasers.

Spoiler for Solution

Spoiler for old wording

• 0

Optical Illusions
Support BrainDen

"To start: Press any key... Where's the 'any' key?" - Homer Simpson

### #2 nzhunter

nzhunter

Newbie

• Members
• 2 posts

Posted 06 May 2007 - 12:45 AM

Hey there

You've got me thinking...

So we came up with 2 different ways to do this, using the scale (we think) only 2x but also 3x too. However, I'm a wee bit confused having read your answer...

Your method states using the scales 3x but in each of the 3 identical steps you state, "Compare 2 groups"... how do you compare 2 groups by using the scales only once?

Logic >>

Step 1: 3 sets of 9 balls, and
Step 2: 3 sets of 3 balls, and
Step 3: 3 sets, 1 ball in each.

Weigh 1 set, you have one weight but you still do not know if you have the heavier ball or if it is with the remaininig 2 sets or 18 balls = you must weigh another of the sets (either 9 balls or the entire 18 balls) to determine which set contains the heavier ball (i.e. you either get an equilibrium or not) >> You used the scale 2x.

Am I missing something here??? It would appear, if I am not incorrect, your method uses the scales 6x???
• 0

### #3 netwebdave

netwebdave

Newbie

• Members
• 4 posts

Posted 06 May 2007 - 04:21 AM

This is talking about the scale that is used to compare two groups. I know I will be banned for using this example but if you ever have seen Celebrity Fit Club, one team stands on one side, the other team on the other and it compares the weight of the two. The heavier group then lowers. It's about the same as a balanced teetor totter with two children. If they both do not move, the heavier side will go to the ground.

So, if you have 3 sets of items, you can weigh two of them. If they are center balanced, then you know the third set is heavier because in this case, one item has to be heavier than the others. Hence, if you had 3 sets of 9 and applied what was stated, you will eliminate 18 objects leaving 9 in one use of the scale. Apply the same theory 2 more times and you are down to 1 object. Hope this helps.
• 0

### #4 nzhunter

nzhunter

Newbie

• Members
• 2 posts

Posted 06 May 2007 - 09:26 PM

Got it now. Cheers.

Will work on our method of using an e-scale only twice!
• 0

### #5 psyber859

psyber859

Newbie

• Members
• 1 posts

Posted 06 June 2007 - 07:39 PM

Actually, if you're very lucky, you might get away with one weighing.

Split the balls into 2 groups of 13, with one ball left out. The heavier group on the scale contains the heavy ball. If both are equal, the odd ball not weighed is the heavy ball.

If you're not lucky, the 13 remaining balls can be broken into different (numeric) groups that still allow you to finish in three steps.
• 0

### #6 akshay.nta

akshay.nta

Newbie

• Members
• 1 posts

Posted 03 July 2007 - 06:23 PM

27 = 9 + 9 + 9
step 1 divide in three equal part and weight two part if both are equal then third one has default ball
if we get default ball part and again divide in three equal part
9 = 3 + 3 + 3
step 2 again do same process step 1
3 = 1 + 1 + 1

step 3 follow the step 1
• 0

### #7 Rudy22

Rudy22

Newbie

• Members
• 1 posts

Posted 10 July 2007 - 04:56 AM

Split the balls into 2 groups of 13, leaving 1 ball unmeasured...if your lucky, the scales will be exactly equal leaving the left out ball the heavier one...so maybe only 1x
• 0

### #8 bigbic

bigbic

Newbie

• Members
• 3 posts

Posted 15 July 2007 - 06:00 AM

nzhunter , your thinking of a normal weight scale, where you put an object(s) on and it gives you the weight of it, the riddle is talking about a balance scale, which has two sides to put weight on to compare them.
• 0

### #9 TorqueRat

TorqueRat

Newbie

• Members
• 3 posts

Posted 17 July 2007 - 01:38 PM

If you are lucky, one weighing is all you'll need. Place one ball on each opposing scale tray and if it tilts, you found the heavy ball. Then discard the other 25.

The puzzle does not specify being able to routinely perform this task.

Yeah, I know, you'll say it is implied.
• 0

### #10 JASON4P8C10

JASON4P8C10

Newbie

• Members
• 7 posts

Posted 28 August 2007 - 04:09 AM

Spoiler for ...

THANK YOU! THANK YOU!
I TAKE DONATIONS

i didnt feel like retyping everything the above is for 1 scale but for "a pair of scales" assuming it isn't like a pair of pants and acutally means 2 scales the minimum weighs is 3
• 0

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

0 members, 0 guests, 0 anonymous users