How Many Pencils?

Date: 09/05/2001 at 21:48:49
From: Thomas
Subject: Math word problem

Al is a commercial artist who uses colored pencils to sketch. How many 
pencils does Al have if all of them are blue except 2, all of them 
are yellow except 2, and all of them are red except 2?

I don't know where to start on this problem.

Date: 09/06/2001 at 14:19:06
From: Doctor Ian
Subject: Re: Math word problem

Hi Thomas,

Let's represent a pencil by the colors that it might have.  For 
example, 

  [r y b]

is a pencil that might be red, yellow, or blue, while 

  [r b]

is a pencil that might be red or blue, but not yellow, and

  [b]

is a pencil that we know has to be blue. 

We know that Al has some pencils, but we don't know how many. Let's 
use '...' to indicate the ones we know are there, but don't want to 
draw:

  [r y b]  [r y b]  [r y b] ... [r y b]

The first sentence tells us that 'all of the pencils are blue except 
2'. What does that mean?  Well, it means that two of them aren't blue, 
and the rest are blue:

  [r y]  [r y]  [b] ... [b]

The next sentence tells us that 'all of them are yellow except 2'.  
Now, none of the blue pencils are yellow, so we can't have more than 
two of them.  So there are three possibilities:

  1. No blue pencils:      [r y]  [r y]

  2. One blue pencil:      [r y]  [r y]  [b]

  3. Two blue pencils      [r y]  [r y]  [b]  [b]

So let's look at those possibilities.  

   1.  [r y]  [r y]

       If it's true that 'all the pencils are yellow, except 2',
       then we're left with two red pencils:

       [r] [r]

       But this makes the next sentence - 'all the pencils
       are red except 2' - false. So we can ignore this 
       possibility. 

   2   [r y]  [r y]  [b]

       If it's true that 'all the pencils are yellow, except 2',
       then one of the [r y] pencils can't be yellow:

       [r y] [r] [b]

       So far so good: All of the pencils are blue, except 2; and 
       all of the pencils are yellow, except 2.  

       What about the next sentence? 'All of the pencils are
       red, except 2'. For that to be true, the [r y] pencil
       can't be red:

       [y] [r] [b]

       All the sentences describe this situation accurately, so
       this looks like a solution to the problem. But we'd better
       check out the final possibility, because sometimes a problem
       can have more than one solution. 

   3.  [r y]  [r y]  [b]  [b]

       If it's true that 'all the pencils are yellow, except 2', 
       then both of the [r y] pencils have to be yellow:

       [y] [y] [b] [b]

       But now there's no way for the next sentence - 'all the 
       pencils are red except 2' - to be true, since we clearly
       have four pencils that aren't red!

So possibilities (1) and (3) don't work out, which makes possibility 
(2) the solution. 

The hardest thing about solving a problem like this is keeping track 
of the things you've already considered. Drawings and tables are 
useful for that, as are trees, and other tools that you'll learn about 
in your later math classes.  

One other thing that's useful is being able to think about problems 
in a very general sense. In this case, the way I _actually_ solved it 
was to think: "The possibilities are that there are no pencils at all, 
1 pencil of each color, or some more complicated answer."  

The case with no pencils can't work, since the '... except 2' part 
will never work. But what about the case with one pencil?  Well, 
suppose I have one red pencil, one blue pencil, and one yellow pencil.  
All but two of them are red, all but two are blue, and all but two are 
yellow. So that is a solution to the problem.

Note that I didn't do any real 'math' in order to get it, apart from 
knowing that 0 and 1 are especially easy numbers to work with. If 1 
hadn't worked, I probably would have tried 2 as well, since it's 
mentioned in the problem. 

Does it seem like cheating to do that? It isn't. Think of it this way, 
suppose I give you a keychain with a bunch of keys, and tell you that 
one of them will get you through a particular door. Would it be 
cheating to try opening the door without even using a key? In fact, 
wouldn't it be silly _not_ to try that? 

A lot of mathematics is really nothing more than trying to find the 
easiest possible way to do something. In that sense, mathematics is a 
clever and constructive kind of laziness. If you approach your math 
classes with that frame of mind, I think it might help you enjoy them 
more than you would otherwise.  

And in fact, here's where the really fun part of the problem begins 
for me.  Having guessed the answer, and having worked it out in kind 
of a clumsy way, I'll probably spend the train ride home tonight 
thinking about whether there is a less clumsy way to work it out, and 
how one might go about making up similar problems. (What if there are 
four colors? What if there are N colors?)

Thanks for asking such an interesting question!

I hope this helps.  Write back if you'd like to talk about this some 
more, or if you have any other questions. 

- Doctor Ian, The Math Forum
  http://mathforum.org/dr.math/