Here are six solutions. Notice that among the six solutions,
there are three different answers. Are they all correct?
Is it possible that this problem has more than one solution?
If not, where did the wrong ones go wrong?
The first three solutions
came all the way from Hong Kong in June 2005:

Hi, I'm Ethan and I'm a sixth grader from Hong Kong International School.
I think the answer for Ballons is 4.
I think this because the data I collected from the notecards state that Anna has two more than Eric,
Hort has five more than David, Eric has less than 20, Hort has more than Anna,
and David has half of Anna's.
Immedietly I knew that David has the least.
So I started with David having 1.
However, the numbers don't fit for everyone. So I tried 2 for David, and it fit for everyone. 
Hi.
We're students in grade 6 from Hong Kong International School,
and we think we have found the answer to Eric's Balloons.
We knew that Dave had half the amount of balloons has Anna.
If Hortensia had 5 more than Dave and more than Anna, we knew that Anna had to have less than 10 balloons.
We therefore figured that Anna had 8 balloons. That would mean Dave had 4 balloons.
If Hortensia had 5 more balloons than Dave, she would than have 9 balloonsmore than Anna.
This would justify that Anna really did have 8 balloons,
which would mean that Eric had 6 baloons2 fewer than Anna.
Thanks for taking time to read this. 
My group figured out the answer to Balloons. Here it is.
Answer: Eric has 4 balloons.
I figured it out by changing all the clues to algebraic equations:
A/2=D
A2=E
D+5=H
H>A
Sincerely,
Janet K

Great! Thanks!
Now, it's not clear how Janet's group went from the algebra to the answer. And how did they use the inequality "H > A"?
The next three solutions may be from the
same class in Clarkstown, Michigan, in Sashabaw (or Sashabae) Middle
School. Thanks to all the group members who submitted!
The first one, where perhaps they stopped
typing a little early...

We figured the problem out by guess and check. We started
with the # 20 and found out that it didn't work then went to 10 and
found that it didn't work then finally we went to 6 and found out
that it did fit with all the clues.
Eric has 4 balloons

The second one, also guess and
check... 
We found the answer by using the guess and check method. We
tried many answers until we came upon the write one. We knew that
Anna had to have an even amount of balloons. We also knew that Eric
couldn't have had more than 20 balloons. Here are our answers:
Eric4
David3
Anna6
Hortensia8

The third is different... 
First we wrote equations for each card
and added them to the chart. Then we realized that Hortensia has to
have 5 more then David. To get Anna's amount then we double David's
amount then we subtracted two from Anna's. Hortensia has 10, Anna has
10, Eric has 8, and David has 5.

Okay, folks, what do you think? We have seen more than one solution. Are they all right? Are there other
solutions? If somebody was wrong, what went wrong? Write to
with your comments!
Now here is a different sort of response, received late in 2008: 
We are a group of eighth grade Fayette County students from West Virginia.
Our first effort to solve "Balloons" was to create a range for each person,
but we soon realized that it wouldn't make sense.
For example, if you chose 5 (out of a range of 28) for Anna,
it wouldn't work since you have to take half of it to find David's amount of balloons,
and you can't have half of a balloon.
So we decided that the best way to find the amount for each person was to make a chart.
name  solution 1  solution 2  solution 3  solution 4 
David  1  2  3  4 
Anna  2  4  6  8 
Hortensia  6  7  8  9 
Eric  0  2  4  6 
Also, we determined that the third response [above] was incorrect because, H must be greater than A.

Okay, what do you think about that? Are all of their solutions solutions?
Did they get them all?
Also: what number patterns do you see? Why do you think they arise? 
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