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Find the Four Integers

Date: 06/10/2003 at 11:51:33
From: Rabbit
Subject: A mathcounts problem.

Four distinct integers a, b, c, and d, have the property that when 
added in pairs the sums 16,19,20,21,22, and 25 are obtained. What are 
the four integers?

Is there any way to do this problem without having to guess and check?
Is there a shortcut?


Date: 06/10/2003 at 13:51:15
From: Doctor Peterson
Subject: Re: A mathcounts problem.

Hi, Rabbit.

Here is a similar, but considerably more complicated, problem:

   Weighing Bales of Hay
   http://mathforum.org/library/drmath/view/56767.html 

We can use similar techniques in your problem. In most such problems 
there will be some guessing and checking, but it is orderly and not 
wasteful.

First, we know that the two smallest numbers must add up to 16, and 
the two largest to 25. We can make an addition table, assuming that 
a<b<c<d:

      a   b   c   d
     --- --- --- ---
  a|  -  16   ?   ?
   |
  b|      -   ?   ?
   |
  c|          -  25
   |
  d|              -

Think about the order of the sums. Clearly a+b < a+c and b+d < c+d; so 
we can indicate relationships this way:

      a   b   c   d
     --- --- --- ---
  a|  -  16 < ? < ?
   |          <   <
  b|      -   ? < ?
   |              <
  c|          -   25
   |
  d|              -

We don't know how b+c and a+d are related; but we know enough to fill 
in a+c and b+d:

      a   b   c   d
     --- --- --- ---
  a|  -  16 < 19< ?
   |          <   <
  b|      -   ? < 22
   |              <
  c|          -   25
   |
  d|              -

This tells us that c = b+3. There are two ways to fill in the last two 
spaces; this tells us that b-a = 1 or 2, and d-c = 1 or 2:

        1   3   2              2   3   1
      a   b   c   d   or     a   b   c   d
     --- --- --- ---        --- --- --- ---
  a|  -  16 < 19< 21     a|  -  16 < 19< 20
   |          <   <       |          <   <
  b|      -   20< 22     b|      -   21< 22
   |              <       |              <
  c|          -   25     c|          -   25
   |                      |
  d|              -      d|              -

But since a+b is even, their difference can't be odd; do you see why? 
Knowing that a+b=16 and b-a=2, we can easily finish the problem.

One thing that might help is to "play" with my ideas before trying to 
work through the logic. That is something I often do when I approach 
a new problem. In this case, that means picking some numbers and 
seeing what sums you get, in order to get a feel for how the sums are 
related to the numbers you start with. The problem involves working 
backward with that idea, so it's a good idea to start by seeing how 
it works forward.

Let's take a set of four numbers at random; say, 6, 8, 11, and 14. We 
can look at all their sums by making an addition table like this:

  +   6   8  11  14
     --- --- --- ---
  6| 12  14  17  20
   |
  8| 14  16  19  22
   |
 11| 17  19  22  25
   |
 14| 20  22  25  28

One thing to notice is that the table is symmetrical; if you draw a 
diagonal from the top left to the bottom right, the numbers "reflect" 
in that "mirror," so that the 14 is opposite another 14, and so on. 
That's because 6+8 = 8+6; we say that addition is commutative. So in 
order to see all possible sums of pairs, we can ignore half of the 
table, which is a duplicate of the other half:

      6   8  11  14
     --- --- --- ---
  6| 12  14  17  20
   |
  8|     16  19  22
   |
 11|         22  25
   |
 14|             28

Notice also that the sums on the diagonal are a number plus itself, 
and we are not allowing that in the problem; so we can remove those:

      6   8  11  14
     --- --- --- ---
  6|  -  14  17  20
   |
  8|      -  19  22
   |
 11|          -  25
   |
 14|              -

Now you can notice some things about the rows. Look at the first row: 
the numbers are in order, because you are adding bigger and bigger 
numbers to 6. The same is true in every row, and in every column. And 
it will always be true. That's why in my answer I put "less than" 
signs into the table, like this:

      6   8  11  14
     --- --- --- ---
  6|  -  14 <17 <20
   |          <   <
  8|      -  19 <22
   |              <
 11|          -  25
   |
 14|              -

This means that

  14 < 17 < 20

  19 < 22

  17 < 19

  20 < 22 < 25

When I worked backward, I knew that these relationships had to be 
true, without having filled in the numbers! And that was the key to 
my solution.

I hope this fills in some background so you can better understand my 
thinking.


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


Date: 06/12/2003 at 11:48:50
From: Rabbit
Subject: A mathcounts problem.

What I don't understand is how you know that b-a = a or 2, and d-c = 1 
or 2 (it was in the first part of the discussion).

Thanks!


Date: 06/12/2003 at 12:03:39
From: Doctor Peterson
Subject: Re: A mathcounts problem.

Hi, Rabbit.

Here is the table I came up with:

      a   b   c   d
     --- --- --- ---
  a|  -  16 < 19< ?
   |          <   <
  b|      -   ? < 22
   |              <
  c|          -   25
   |
  d|              -

I said that the difference from 16 to 19 tells us that c = b+3. Then 
there are two numbers left to fill in the two missing spaces, which 
can be done in two ways:

      a   b   c   d   or     a   b   c   d
     --- --- --- ---        --- --- --- ---
  a|  -  16 < 19< 21     a|  -  16 < 19< 20
   |          <   <       |          <   <
  b|      -   20< 22     b|      -   21< 22
   |              <       |              <
  c|          -   25     c|          -   25
   |                      |
  d|              -      d|              -

The first of these shows that changing from a to b (going from the 
first to the second row) adds 1 (19 to 20, 21 to 22). So b must be 1 
more than a. Similarly, the second goes from 19 to 21 and 20 to 22, 
so b must be 2 more than a. You can do the same with the last two 
columns to show that d is 1 or 2 more than c.

Does that help?

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


Date: 06/12/2003 at 12:09:38
From: Rabbit
Subject: Thank you (A mathcounts problem.)

Now I understand. Thank you so much for taking the time to 
help me, Doctor Peterson!
Associated Topics:
High School Puzzles

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