## 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 in a Harder Way

Best Answer bonanova, 18 January 2008 - 06:49 PM

Jkyle1980's solution is correct, and much less verbose than what follows.

You can distinguish among 3**N cases in N weighings.
There are 54 possible cases in this puzzle [one of 27 coins is heavy or light].
So three weighings [27 cases] won't do it, but four [81 cases] can.

Spoiler for Here's a very slightly modified approach
Go to the full post

83 replies to this topic

### #1 wangsacl

wangsacl

Newbie

• Members
• 2 posts

Posted 18 June 2007 - 05:03 AM

You've got 27 coin, each of them is 10 g, except for 1. The 1 different coins has a masses different. It could be 9 g, or 11 g (heavier, or lighter by 1 g).
The question is, in how many minimum ways, you can determine the different coins, and is it heavier, or lighter?
I have no answer to this question.
Do you?
• 0

### #2 mdsl

mdsl

Junior Member

• Members
• 51 posts

Posted 18 June 2007 - 01:53 PM

Assuming you have a simple scale, my answer is 9.

First divide the coins into 3 piles of 9.

Measure 2 piles at a time until you find the pair that is equal in mass. This can take up to 3 tries, but if you were lucky you could get it on the first try.

Put the piles of coins with equal mass to the side.

Repeat the process this time with 3 piles of 3 coins, maximum 3 measurements.

Repeat the process this time with the 3 remaining coins, find the two that are equal, the remaining coin has a different mass.

If you are a very luck person you coulp potentially only have to make 4 meaurements, 3 to rule out equal pairs and 1 to compare the final coin to the rest.

9 is the most you would have to do the efficient way, and at any point you can observe is the unequal pile has more or less mass.
• 0

### #3 CedricLi

CedricLi

Newbie

• Members
• 5 posts

Posted 16 January 2008 - 12:28 PM

I guess the question is lack of some information such as, by using a scale with weight of 1g, 2g, 5g, and 10g, isn't it?
• 0

### #4 bonanova

bonanova

bonanova

• Moderator
• 6161 posts
• Gender:Male
• Location:New York

Posted 16 January 2008 - 05:14 PM

I guess the question is lack of some information such as, by using a scale with weight of 1g, 2g, 5g, and 10g, isn't it?

It's a balance scale that compares what's in the two pans.
You can get the answer by just comparing groups of coins.
• 0

Vidi vici veni.

### #5 bonanova

bonanova

bonanova

• Moderator
• 6161 posts
• Gender:Male
• Location:New York

Posted 16 January 2008 - 07:21 PM

Duplicate post.
Looking around the forum for an eraser ... <_<
• 0

Vidi vici veni.

### #6 PDR

PDR

Junior Member

• Members
• 50 posts

Posted 18 January 2008 - 01:05 AM

Spoiler for fewer than 9...

• 0

### #7 PDR

PDR

Junior Member

• Members
• 50 posts

Posted 18 January 2008 - 01:05 AM

Spoiler for fewer than 9...

• 0

### #8 PDR

PDR

Junior Member

• Members
• 50 posts

Posted 18 January 2008 - 01:28 AM

sorry - accidentally hit submit - here's my answer...
Spoiler for fewer than 9...

• 0

### #9 Jkyle1980

Jkyle1980

• Members
• 157 posts

Posted 18 January 2008 - 04:24 AM

I think I've got it in 4 measurements:

Separate the coins into 3 stacks of 9 (A, B, C). Weigh stack A against B and then A against C. Take the stack with the different weight (note lighter or heavier) and break it into 3 stacks of 3 (D, E, F). Weigh stack D against E. If D and E are equal, then F is the odd stack. If D and E are not equal, the lighter or heavier (based on the A, B, C comparison) is the odd stack. You now have three coins (G, H, I). Weigh G and H. If G equals H, then I is the odd and is lighter or heavier (based on the A, B, C comparison). If G and H are not equal, then the lighter or heavier (based on the A, B, C comparison) is the odd coin.
• 0

### #10 bonanova

bonanova

bonanova

• Moderator
• 6161 posts
• Gender:Male
• Location:New York

Posted 18 January 2008 - 06:49 PM   Best Answer

Jkyle1980's solution is correct, and much less verbose than what follows.

You can distinguish among 3**N cases in N weighings.
There are 54 possible cases in this puzzle [one of 27 coins is heavy or light].
So three weighings [27 cases] won't do it, but four [81 cases] can.
Spoiler for Here's a very slightly modified approach

• 0

Vidi vici veni.

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

0 members, 0 guests, 0 anonymous users