## 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

# Selections and weighings

2 replies to this topic

### #1 spoxjox

spoxjox

Junior Member

• Members
• 97 posts

Posted 13 December 2007 - 08:33 PM

You have 81 black balls and 19 white balls, of identical shape, weight, and feel, in a barrel. You select balls blindfolded.

How many balls must you select to be assured of selecting a black ball?
Spoiler for solution

How many must you select to be assured of getting a white ball?
Spoiler for solution

How many must you select to be assured of getting all the black balls?
Spoiler for solution

Now that you have all 81 black balls, you are told that they are not truly identical, but that one is slightly heavier than the others. You are given a set of balance scales and instructed to find the heavy ball. What is the minimum number of weighings in which you can guarantee to find the heavy ball?
Spoiler for solution

Using the above method, what is the *minimum* number of weighings in which you could actually find the heavy ball?
Spoiler for solution

Eighteen of the nineteen white balls are identical. The nineteenth is of a slightly different weight, but you don't know if it's heavier or lighter (and can't tell by feel). Using the balance scales, what is your weighing strategy for finding the odd ball and what is the minimum number of weighings to guarantee finding it? Assume that the eighteen balls are lettered from A to R so that you can tell them apart. (I don't have the optimal solution to this, since I just made it up and it's not obvious to me, but I'll think about it while reading other's responses.)
• 0

### #2 spoxjox

spoxjox

Junior Member

• Members
• 97 posts

Posted 13 December 2007 - 09:39 PM

EDIT: I just saw that I violated my own conditions...oops. I solved the problem for eighteen balls instead of nineteen! Well, here is my "solution" anyway, for whatever it's worth. The real solution will be similar, I think.
******************
What a ripoff! I just wrote up a solution for the last problem that identified the odd ball in three or four weighings, but I wasn't allowed to post it, and now it's lost! Let me try to reproduce it.
Spoiler for solution

*******************
EDIT: Maybe this will do it.
Spoiler for solution

• 0

### #3 bonanova

bonanova

bonanova

• Moderator
• 5918 posts
• Gender:Male
• Location:New York

Posted 14 December 2007 - 07:45 AM

20, 82, 100, 4.
There are exactly 81 possibilities.
Each weighing has 3 distinguishable outcomes.
Four weighings can distinguish among 3x3x3x3 = 81 possibilities.
Devise a 1st weighing that reduces the number of possibilities to 27 -> 27 against 27; with the 27 suspect balls,
Devise a 2nd weighing that reduces the number of possibilities to 9 -> 9 against 9; with the 9 suspect balls,
Devise a 3rd weighing that reduces the number of possibilities to 3 -> 3 against 3; with the 3 suspect balls
The 4th weighing is 1 against 1; it determines the heavy ball.
Four is the fewest weighings, using this method.
4.
Three weighings can distinguish among no more than 3x3x3=27 possbilities.
With 19 balls that could be heavy or light you have 19x2=38 possibilities.
Three weighings won't do it.
Use the above method.
• 0
The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
- Bertrand Russell

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

0 members, 0 guests, 0 anonymous users