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Numbers on a Star Brain Teaser

```Date: 01/03/2008 at 22:16:17
From: Linda
Subject: 5th grade brain teaser - Mom, Dad &amp; older sister can't solve

We have a number sequence of 20, 21, 22, 23, 24, 25, 27, 28, 29 and
31.  The numbers are to be placed on a star so that each line of four
numbers has a sum of 100.  Each number can only be used once.

The instructions advise my 5th grade daughter to "guess and check".
My husband, 14-year old daughter and me (mom) have tried the guess and
check method for about an hour, trying all sorts of combinations.
Obviously we all are missing something that is needed to solve the
problem.

We've tried listing all number sequences that add up to 100.  Then we
assign numbers to the outside points of the star and use a Sudoku-type
method of figuring out what the numbers would be on the inside points
of the star.  But we're all scratching our heads on this, this is
going to be very embarrassing if this is easy.  Please help, thank you
so much.

```

```
Date: 01/04/2008 at 10:39:52
From: Doctor Rick
Subject: Re: 5th grade brain teaser - Mom, Dad &amp; older sister
can't solve

Hi, Linda.

I don't have any good suggestions beyond what you have done.

Well, one minor thing: you can reduce the size of the numbers you have
to work with, by subtracting 20 from each number.  Then the row-sums
are reduced by 4*20 = 80, so we're looking for ways to put the numbers
0, 1, 2, 3, 4, 5, 7, 8, 9, and 11 into the star so the row-sums are
20.

I too listed all the sets of 4 of these numbers whose sum is 20; I
found 13 sets (ignoring order).  But it looks like too much work.

My instinct, when I can't do much better than guess-and-check and
there are a lot of choices to check, is to write a computer program.
I did that and found 120 solutions. However, for each distinct
solution, there are five ways to rotate the star and another five
ways with the star flipped around, so the solutions come in groups
of 10.  There are 12 distinct solutions, when rotating or flipping the
star is not considered to produce a different solution.

Note, though, that there are 10! = 3,628,800 ways to arrange the 10
numbers.  Taking out the factor of 10 for rotations and flips, there
are still 362,880 different stars, of which only 12 are solutions to
the problem.  I don't know how anyone can be expected to find one of
them by guess-and-check, without a lot of luck!

Here is the first solution my program found:

24

20    29    28   23

22      21
31
25          27

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

```

```
Date: 01/04/2008 at 21:34:24
From: Doctor Peterson
Subject: Re: 5th grade brain teaser - Mom, Dad &amp; older sister
can't solve

Hi, Linda.

I was going to write an answer almost exactly like the first part of
Dr. Rick's answer last night, just suggesting you reduce the size of
the numbers and telling you that these problems are inherently hard,
so not to worry about it.  (I've struggled for a long time with some
very similar ones.)  But I left it in case anyone could give a more
positive answer than that.

Then I continued to think about the problem as I brushed my teeth, and
within a few minutes I had a solution with remarkably little work.  I
don't know that a child could be expected to do this (since it's a lot
more than mere guess-and-check), but here's an outline of my method:

1. Subtract 20 from each number, and thus subtract 80 from the
expected sum of each row; this can be verified by noting that the sum
of all 5 rows will be twice the sum of the 10 numbers, so the sum of
one row is 2*50/5 = 20.  Now our goal is to place 0, 1, 2, 3, 4, 5, 7,
8, 9, and 11 in the star so that each row sums to 20.

2. Place the largest number, 11, somewhere; I put it at the top.  That
number is in two rows; the sum of the remaining 3 numbers in each row
has to be 20-11 = 9.  List the ways to make 9 from three of the numbers.

11
/ \
/   \
x------a-----d------y
\  /       \  /
b         e
/ \     / \
/     z     \
/  /     \  \
c             f

3. Pair up those five sets of three to find two that do not overlap,
since they can't share a number.  This gives only two such pairs,
namely 0+1+8=2+3+4 and 0+2+7=1+3+5.

4. For each of these, do the following: Consider that the remaining
three rows each contain one number from each list (a+d, b+f, and c+e
as shown above), combined with two of the remaining three numbers
(shown as xyz above).  List the three sums formed by pairs of the
latter numbers, and from those find the sums that the pairs of numbers
from the two sets of three have to add to.  For the second pair of
sets, the remaining numbers are 4, 8, and 9, giving sums 12, 13, and
17, so that pairs from the two sets have to add to 8, 7, and 3.  (The
first pair of sets of three don't work; you can't do this step.)

5. Find pairs that give these three sums; there is exactly one way to
do this:

0 + 3 = 3 (will be in row with 8+9)
2 + 5 = 7 (will be in row with 4+9)
7 + 1 = 8 (will be in row with 4+8)
--- ---
9   9

6. Now place 0, 2, 7 on one row emanating from the 11 (my abc above),
and 3, 5, 1 on the other (my def), making sure that the correct pairs
lie on rows crossing them.

7. Finally, place 4, 8, and 9 in the spots (xyz) where they give the
right sums.

8. Now add 20 to the numbers, and you're done!

I think all the solutions are equivalent in the sense that the five
rows always contain the same five sets of numbers.

Again, don't let anyone feel bad at not getting the answer by
guess-and-check; brain teasers are fun when you solve them, but even
the best of us don't always do so.

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