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### Thoughtful Trial and Error

```Date: 05/21/2002 at 23:42:17
From: Ashley
Subject: It will make you dizzy!

Dear Dr. Math,

Here is the problem I need to figure out:

O
O O
O O O
O O O O

Place the numbers 1 through 10 in the circles so that:

a) the rows of three circles have the same sum; and

b) the rows of four circles have the same sum (not necessarily the
same sum as the rows of three circles).

I have tried trial and error for hours on this annoying problem. Is

Thanks,
Ashley
```

```
Date: 05/22/2002 at 23:06:15
From: Doctor Peterson
Subject: Re: It will make you dizzy!

Hi, Ashley.

To tell the truth, this kind of puzzle is largely trial and error, so
it's hard to give hints without giving the answer. But I just sat down
and solved it in a few minutes, using what you might call experienced
trial and error. I'd like to show you some of what I did.

One trick I know for puzzles like this is to think about what happens
if you add up all the rows together. In this case, if you add the
three rows of four (whose sums I'll call A) and the three rows of
three (whose sums are B), you have added in each number twice, except
that the middle one is counted three times. But the sum of the
numbers from 1 to 10 is 55. So

3A + 3B = 2*55 + X

where X is the middle number. That limits what X can be: since 2*55 =
110 is one less than a multiple of 3, X must be one more than a
multiple of 3: 1, 4, 7, or 10.

So now let's choose a middle number; I picked 10 to try first. (Since
there's no number right in the middle between 1 and 10, an extreme
seemed like a good choice!) Then

3(A + B) = 110 + 10

A + B = 40

I just chose a neat set of numbers for the three corners; taking the
four used numbers out of the set, the sum of the rest of the numbers
is 55 minus their sum. A third of that sum tells me what the sum of
opposite numbers has to be; and I could tell from the sums of the
pairs of numbers at the corners, what sums I needed for the middle
two numbers along each side (that is, three different sums). Then I
just tried placing the six remaining numbers in their places (with
the corner numbers _not_ yet written) to fit those requirments, and
then I could tell where the corner numbers had to be. My first choice
turned out to work; that suggests that there may well be other
solutions. I know there will be a solution with 1 in the middle
(obtained from my solution by subtracting each number from 11). There
may be still more.

Have fun trying for your own solution! I hope this gives you some
good ideas, without taking away the fun of discovery, and a lesson

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

```
Date: 05/23/2002 at 08:53:13
From: Doctor Peterson
Subject: Re: It will make you dizzy!

Hi again, Ashley.

When I picked up my papers last night I looked at my solution and
realized it was wrong. In case that misleads you, I want to let you
know that I did find a solution, but it used 7 in the middle rather
than 10; I think I'm convinced that 10 in the middle doesn't work.

Also, when I found the real solution I used a different trick than I
had mentioned. Since I had determined the sum A+B for a given middle
number X, I realized that I could subtract that sum from 55 and find
the sum of the three numbers at each corner. Then I could find all
the ways to divide the remaining numbers into three groups with that
sum (there were only two ways!), and then try each of the three
possible placements of one of those groups and see whether it led to
a solution. That's how I convinced myself there is no solution with
X=10.

The puzzle is harder than I thought, but with careful trial and error
it can be done!

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

```
Date: 05/25/2002 at 19:52:17
From: Ashley
Subject: It will make you dizzy!

Question submitted via WWW:
Hey Dr. Peterson,
Thanks so much. I finally figured it out! Yay!
Thanks again,
Ashley
```
Associated Topics:
High School Basic Algebra
High School Puzzles
Middle School Algebra
Middle School Puzzles

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