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Number Puzzle


Date: 02/12/97 at 11:22:02
From: marko kuo
Subject: problem solving

I was given a number puzzle to solve in my problem solving class and
I kind of got stuck. I suspect there might be a theory behind it. The 
problem is to find a number with 9 digits in which:

      the first digit is the number of 0's in this number
      the second digit is the number of 1's in this number
      .
      .
      the 9th digit is the number of 8's in this number

Thanks,
Kuo


Date: 02/13/97 at 15:39:35
From: Doctor Wilkinson
Subject: Re: problem solving

This is really quite tricky.  Don't feel bad about getting stuck. I 
got stuck a couple of times myself.  

What I did was write a heading for each digit:

        0 1 2 3 4 5 6 7 8

I then started filling in possibilities to try to narrow things down.

Suppose we had:  0 1 2 3 4 5 6 7 8
                                 2

Then we need to have an 8 in two places, but that gives us 8 of two
different digits, and we only have 9 digits.  So under the 8 we must 
have a 0 or a 1, and similarly for 5, 6, and 7.

Now let's try:  0 1 2 3 4 5 6 7 8
                                1

Now we've got to put that 8 somewhere, which means that there's 8
of something, and there just isn't room, because there are only
7 places left.

So we must have:  0 1 2 3 4 5 6 7 8
                                  0

Moving on to the 7, suppose we had:
  
                 0 1 2 3 4 5 6 7 8
                               1 0

Then that 7 has to go someplace, and we also have a 1 already, and
again there isn't room.

So we have:  0 1 2 3 4 5 6 7 8
                           0 0

And let's suppose we have a 1 under the 6.  This gives us:

        0 1 2 3 4 5 6 7 8
                    1 0 0

Now there isn't room for 6 of anything except 0, so we have to have:

        0 1 2 3 4 5 6 7 8
        6           1 0 0

We then have to put the 1 someplace.  It can't go under the 1, because
that would give us two 1's.  But if it doesn't go under the 1, then we
have to have a 2, 3, 4, or 5, and once again we don't have room.

So we have:  0 1 2 3 4 5 6 7 8
                         0 0 0

Let's try:   0 1 2 3 4 5 6 7 8
                       1 0 0 0

Again, you have to put the 5 under the 0:

        0 1 2 3 4 5 6 7 8
        5         1 0 0 0

Now you can't put the 1 under the 1, just like last time.  So let's 
put a 2 under the 1 and see what happens:

        0 1 2 3 4 5 6 7 8
        5 2       1 0 0 0

Now we have a 2, so we need to put something other than 0 under the 2,
which leaves us no choice but to put the two more 0's we need under 
the 3 and the 4, and there's only room for one 2, which gives us:

        0 1 2 3 4 5 6 7 8
        5 2 1 0 0 1 0 0 0

It works!

You should finish this by eliminating any other possibilities.

-Doctor Wilkinson,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   
    
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