Solving Problems by Making Organized Lists

Date: 04/09/2008 at 18:36:16
From: Remy
Subject: not enough room to explain

this is a problem on my math homework. it is the hardest one on my
page and i did all the others ones and my parents cant even help me. 
so here is the problem.  as Jamey bounces on the trampoline 11 coins
fall out of his pocket totaling 1.37. what are these coins? what about
14 coins and 2.10?

i thought maybe starting out with quarters but that did not work so i
knew it had to be a lot of different type of coins. i started doing 11
divided by 137 but then the period ended so i could not get to it
until 5 minutes ago.

Date: 04/10/2008 at 09:44:08
From: Doctor Ian
Subject: Re: not enough room to explain

Hi Remy,

You need 11 coins that add up to 137 cents.  To finish with 7 cents,
there are two possibilities:

 1) 2 of them are pennies. 
    In that case, you need to find 9 coins that add up to 135 cents.

 2) 7 of them are pennies.
    In that case, you need to find 4 coins that add up to 130 cents.

So now you have two smaller problems that you can work with.  We can
do the same thing again, with quarters. 

  1) Find 9 coins that add up to 135 cents.

     1a) If 1 is a quarter, 
         find 8 coins that add up to 110 cents.

     1b) If 2 are quarters,
         find 7 coins that add up to 85 cents.

     1c) If 3 are quarters,
         find 6 coins that add up to 60 cents

     1d) If 4 are quarters, 
         find 5 coins that add up to 35 cents. 

     1e) If 5 are quarters,
         find 4 coins that add up to 10 cents.

Note that, as an added bonus, since I've already considered pennies
and quarters, I don't even have to think about them when looking at
(1a) through (1e).  I could really write them as 

  1) Find 9 quarters, dimes, and nickels that add up to 135 cents.

     1a) If 1 is a quarter, 
         find 8 dimes and nickels that add up to 110 cents.

     1b) If 2 are quarters,
         find 7 dimes and nickels that add up to 85 cents.

     1c) If 3 are quarters,
         find 6 dimes and nickels that add up to 60 cents

     1d) If 4 are quarters, 
         find 5 dimes and nickels that add up to 35 cents. 

     1e) If 5 are quarters,
         find 4 dimes and nickels that add up to 10 cents.

And when I write it this way, it reminds me that I have to consider
the possibility that there are NO quarters:

  1) Find 9 quarters, dimes, and nickels that add up to 135 cents.

     1a) If 0 are quarters,
         find 9 dimes and nickels that add up to 135 cents.

     1b) If 1 is a quarter, 
         find 8 dimes and nickels that add up to 110 cents.

     1c) If 2 are quarters,
         find 7 dimes and nickels that add up to 85 cents.

     1d) If 3 are quarters,
         find 6 dimes and nickels that add up to 60 cents

     1e) If 4 are quarters, 
         find 5 dimes and nickels that add up to 35 cents. 

     1f) If 5 are quarters,
         find 4 dimes and nickels that add up to 10 cents.


And now I can see how to answer the question!  Because I know how to
get 6 coins that add up to 60 cents.  I just use 6 dimes.  Working
backwards,

    6 dimes give me    60 cents.
    3 quarters give me 75 cents.
    2 pennies give me   2 cents.
                      ---
                      137 cents

But if I didn't see that, I could keep going down that path:

 1d) If 3 are quarters,
     find 6 dimes and nickels that add up to 60 cents

     1c1) If 0 are dimes, 
          find 6 nickels that add up to 60 cents.

     1c2) If 1 is a dime,
          find 5 nickels that add up to 50 cents.

     1c3) If 2 are dimes,
          find 4 nickels that add up to 40 cents.

     1c4) If 3 are dimes,
          find 3 nickels that add up to 30 cents.

     1c5) If 4 are dimes,
          find 2 nickels that add up to 20 cents.

     1c5) If 5 are dimes,
          find 1 nickel that adds up to 10 cents.

     1c5) If 6 are dimes,
          find 0 nickels that add up to 0 cents.   

Here, (1c5) works out, and the others are so obviously wrong that I
can cross them off without having to work very hard.  But if none of
these worked out, I would know that the answer had to be down one of
the other paths. 

Now, the important thing here isn't to answer the question.  No one
cares about the answer.  What's interesting here is that I could
systematically change each problem into a set of smaller problems,
until I get one so small I can solve it in my head. 

It's kind of like if you've lost your watch, and you're looking for it
in your house.  You could start looking around randomly, or you could
formulate a search plan.  For example,

  It's got to be in the basement, or on the first floor, or on
  the second floor:

  1) basement

  2) first floor

  3) second floor

If it's on the second floor, it's got to be in one of the rooms there:

  1) basement

  2) first floor

  3) second floor
     3a)  My room
     3b)  Pat's room
     3c)  Mom and Dad's room
     3d)  The bathroom
     
If it's in my room, it's got to be in one of a few places:

  1) basement

  2) first floor

  3) second floor
     3a)  My room
          3a1) In or on the dresser
          3a2) In the closet
          3a3) On the floor, behind or under some furniture
          3a4) In the bed
           .
           .
     3b)  Pat's room
     3c)  Mom and Dad's room
     3d)  The bathroom

The nice thing about this is that, if the watch is somewhere, you WILL
find it (assuming you've been careful about constructing your list);
and you won't ever look in the same place more than once. 

A lot of math problems are basically the same way.  You start with the
original problem, and say:  Well, the answer has to be down one of
these paths.  And then you WRITE DOWN THE PATHS!  That's the crucial
part, that people tend to forget.  Because when you write down what
you're doing, it helps you (1) figure out the next thing to do, and
(2) avoid doing the same things over and over again.

And then for each path you've started, YOU DO THE SAME THING AGAIN,
i.e., you look at the paths that extend the ones you've started.  And
again, you write it all down. 

Note that sometimes a problem like this won't be to find ONE way to
get the right amount of money, but to find ALL ways to get it.  (In
the case of searching the house, maybe you've lost two earrings, or
five baseball cards, or whatever, and you need to find all of them.)

Also, sometimes a problem is to show that something CAN'T be true. 
(In the case of searching the house, maybe a friend thinks he might
have left something valuable at your house, and you have to show that
this isn't the case.)

In either of these cases, you have to keep going until you've examined
all of the possibilities.  So a list isn't just convenient, it's
indispensable. 

Does this make sense?  Try doing this with your second problem, and
let me know how that goes, okay? 

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