Find the Seven Numbers

Date: 06/07/2003 at 21:18:57
From: Jane
Subject: Puzzles

There are seven numbers, A, B, C, D, E, F, and G.

When all these numbers except A are added together, 116 is obtained.
When all these numbers except B are added together, 122 is obtained
When all these numbers except C are added together, 123 is obtained
When all these numbers except D are added together, 108 is obtained
When all these numbers except E are added together, 110 is obtained
When all these numbers except F are added together, 119 is obtained
When all these numbers except G are added together, 100 is obtained

Find each of the seven numbers.

Date: 06/08/2003 at 13:10:34
From: Doctor Achilles
Subject: Re: Puzzles

Hi Jane,

Thanks for writing to Dr. Math.

Let's start with C. You know that C is the smallest because you get 
all the way to 123 when you add the others.

What is the second-smallest number? B, right? When you add all the 
numbers but B, you get 122, so B is second-smallest.

So we know C is smallest, and B is second-smallest, but how much 
greater than C is B?

In one case, we have:

  A + C + D + E + F + G = 122

and in the other we have:

  A + B + D + E + F + G = 123

Substituting B for C gets us 1 more, so B must be 1 greater than C.

We can say that mathematically like this:

  B = C + 1

Since B is exactly equal to C+1, later on, whenever we see a B, we 
are free to substitute C+1 for it if that helps.

Let's write that down and remember it; it will be important later.


What is the third-smallest number?  It has to be F, right?

How much bigger is F than B?  Well we know:

  A + B + C + D + E + G = 119

and in the other we have:

  A + C + D + E + F + G = 122

So by substituting F for B we get 3 more, and F must be 3 greater than 
B.

Or in math notation:

  F = B + 3

We can even go a little farther. Because B=C+1, we can substitute C+1 
in that equation:

  F = C + 1 + 3

Or

  F = C + 4

Let's write that down and remember it.


What about the next-smallest number?  That has to be A, right?

With the same math tricks I used for F, we can find that

  A = F + 3

[you should double-check my work here, because I didn't write it out]

And again, we know that F=C+4, so we can substitute C+4 for F and get:

  A = C + 4 + 3

or 

  A = C + 7


Next we can figure out that:

  E = A + 6

So

  E = C + 7 + 6

Or

  E = C + 13

And we can also find out that:

  B = C + 15

  G = C + 23

So let's review what we know:

  A = C + 7

  B = C + 1

  C = C

  D = C + 15

  E = C + 13

  F = C + 4

  G = C + 23

We also know that:

>When all these numbers except G are added together, 100 is obtained

Or in math talk:

  A + B + C + D + E + F = 100

Now, we know that A=C+7, which means we can substitute C+7 for A:

  C + 7 + B + C + D + E + F = 100

Or with a bit of rearrangement:

  B + C + C + D + E + F + 7 = 100

C+C equals 2C, so we can also write this:

  B + 2C + D + E + F + 7 = 100

Now, we know that B=C+1, so we can substitute that in:

  C + 1 + 2C + D + E + F + 7 = 100

Which we can rearrange:

  C + 2C + D + E + F + 7 + 1 = 100

Or

  3C + D + E + F + 8 = 100

We also know that D=C+15, so we can substitute that in:

  3C + C + 15 + E + F + 8 = 100

Which we can re-write:

  4C + E + F + 8 + 15 = 100

Or

  4C + E + F + 23 = 100

We can substitute C+13 for E. Then after we rearrange, we can 
substitute C+4 for F. Then we can solve for C just as in a regular 
algebra problem. Once we know C, we can go back to our list of 
equations:

  A = C + 7

  B = C + 1

  C = C

  D = C + 15

  E = C + 13

  F = C + 4

  G = C + 23

to find each of the other variables.

I hope this helps.  If you have other questions or you'd like to talk 
about this some more, please write back.

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

Date: 04/23/2007 at 11:56:53
From: John
Subject:  Simpler solution technique to "Find the Seven Numbers"

I'd like to point out that the symmetry of this problem allows it to be
solved much more simply.  Writing the equations this way,

      B + C + D + E + F + G = 116        (1)
  A     + C + D + E + F + G = 122        (2)
  A + B     + D + E + F + G = 123        (3)
  A + B + C     + E + F + G = 108        (4)
  A + B + C + D     + F + G = 110        (5)
  A + B + C + D + E     + G = 119        (6)
  A + B + C + D + E + F     = 100        (7)

we can see that exactly one unknown is missing from each of the equations.  
So adding all the equations together, gathering like terms and then dividing
both sides by the number of equations (6) gives us

  6(A + B + C + D + E + F + G) = 798

    A + B + C + D + E + F + G  = 133     (8)

Now we can subtract each of the original equations from this, to get the value
of the missing unknown, e.g., subtracting (3) from (8) gives

    A + B + C + D + E + F + G  = 133            (8)

    A + B     + D + E + F + G  = 123            (3)
 
            C                  = 133 - 123

                               = 10

and so on for the rest of the equations. 

The method used by Dr. Achilles method may be more general, but it is a lot
more work, which carries with it a larger potential for making a mistake. 
The solution above is a good demonstration of how much effort can be saved
through the appropriate use of symmetry.

-John