Eighth graders in Monroe City, MO
A second group in Monroe City, MO
responds
Pamela (Brake Middle School) disagrees with the
first response, and then corrects her own...
Work on the limit from some seventh
graders in Indiana
Work on the limit from some seventh graders in Maine

Here is some brand new work on the limit from some students; I don't know where they're from! 
Dear Tim,
Hi. We deleted this mesage about 5 times but here it is. We found a pattern.
You have your basic numbers 112 using only the letters A & P. then, the
numbers 1324 you start using the letter E .At twentyfive you do E+1=P=EP=26
and 27=E+2=PP=EPP and so on until 63. At 63 You take 125  63 that equals 62
so you do Geeaapp. This is 125  63=62 the eeaapp is 62. After 63 for 64 you
have 12564=63 or Geeaap and so on until you reach 125 which is just G. Then
for 126 you do G+1=P=GP=126. For 127 you do G+2=PP=GPP=127. You follow this
pattern as we explained up to 250. Two fifty is GG and 251 is GG+1=P=GGP=251
. You follow this pattern and you find that you cannot do 313. From Jen, Tim,
Leah, Andrea & Robby

Here's
a solution from a group of eighth graders at Monroe City Middle School in Monroe
City, MO:

Megan, Jennifer, Ashley, Joslynn, and Matthew of Monroe City,
MO thought that this problem, GEAP!, was a little tougher than
Znorlian. It took us about thirty minutes to complete this problem.
Here is what we came up with:
G= 125 g= 125
E= 25 e= 25
A= 5 a= 5
P= 1 p= 1
The first clue told us A= 5, AA= 10. The second clue told us
PA= 6 and pA= 4. And since A was 5, P had to be 1 and p=1 because
clue third says lower case letters are negative and upper case
letters are positive. The third clue also says GAA= 135, AA= 10, so
G= 125. It also says no letter can be used more than twice in a
number. Clue four says it doesn't matter what order the letters are
in and that Ep= 24, since p= 1, E has to equal 25. The only way to
make 13 is AApAp, but you can't do that because A is used more than
twice. Here is a list of numbers up to 12:
1= P
2= PP
3= pAp
4= pA
5= A
6= AP
7= APP
8= pApA
9= AAp
10=AA
11=AAP
12=AAPP
teacher: Mrs. Barbara Carson, Monroe City Middle School [1997, I think]

eeps comments: impressive work. Do people
out there agree with these kids in Missouri? Do the
numbers  1, 2, 25, 125  ring any bells for you?
Also: the second part of the problem is deciding
what the smallest number is that you can't make in
GEAP. A different group of students (from the same class)
wrote about that... (on the left)
...and on the right, a group of students from Hermann Missouri find a typo.

The Eighth Grade Monroe City
Middle School R.E.A.C.H. class including Kimberly, Micah, Erica,
Allie, and Tina, found a solution for Geap. We decided that if you
get a solution for the numbers 110, 20, 30, 40, 50, 60, 70, 80, 90,
100, 200, and 300, then you can find a solution for the numbers
between 1 and 300. Since GGEEAAPP is 312, then it is impossible to
represent a number higher than 312, so the answer must be 313. Here
is how we got that:
1P 50EE
2PP 60EEAA
3App 70AA
4Ap 80GeeA
5A 90Geaa
6AP 100Ge
7APP 200GGee
8AApp 300GGEE
9AAp
10AA G=125
20Ea E=50
30EA A=5
40EEaa P=1
teacher: Mrs. Barbara Carson, Monroe City Middle
School [1998]

Dear Tim,
We are a group of enriched 6th grade students at Hermann Middle School in MO.
We've discovered a small typographical error
in the following question you've posted on your website:
Do the numbers  1, 2, 25, 125  ring any bells for you?
Where you listed "1, 2, 25, 125,"
the number 2 should actually be 5.
With this change, the numbers are in a base 5 system.
We figured this out by realizing that 25 times 5 = 125,
but 1 times 5 is not 2. That's when we realized the mistake.
Sincerely,
Emma, Jamie, and Andrew
(2008)


eeps comments: Cool ! Interesting reasoning.
Anybody out there want to refute it?
Meanwhile, Pamela, from
Brake Middle School disagrees with the first answer about 13 being
the smallest unmakeable counting number. I especially like that she
showed how to make 13, which is a great proof! What do you think of
HER reasoning?

The eighth grade students Monroe County have come up with the
wrong solution. Here is my answer.
The question was, "What is the smallest counting number that
you cannot represent on Geap?" They said that the answer is 13 and
that the only way to reach 13 was with AApAp. But that is not true.
My answer is 187. I say this because, you can also do Eaapp to reach
13. This is my solution: GGEEAAPP=312. So I figure if you add in the
biggest  [that is, negative] number you will come up with the
smallest counting number you cannot get. After adding g my new
solution is GGgEEAAPP. This = 187. The answer is 187.
Sincerely,Pamela

I wrote back to Pamela, inquiring about her
reasoning and wondering where (geographically) she was. She
replied:

This is Pamela. I am an eighth grader at
Brake Middle School in Taylor, Michigan. I would like to withdraw my
previous submission and submitt a new one.
After writing you I thought long and
hard about my response and it occured to me that I was wrong. But I
do have the correct answer now. It is GGEEAAPPP. This =313. It is
this because GGEEAAPP=312. So if you add one to 312 you get the
smallest counting number that can not be represented on
GEAP.
So I say I'm wrong with my first answer
because if that were the way it is to be done that you could say 1 is
the smallest counting number because pPP=1. So my final answer is
313.
Thank You ,
Pamela

Comments on the
limit from Amy Goff's 7th grade math applications class in
Cloverdale, Indiana...

Rachel Jones: I believe that every number can be made until you run out of
letters. You can use each letter twice, so you would put G,E,A,and P twice.
Then 1 number above that is impossible to make correctly in the geap
method. All the letters put twice is equal to 312. So 312 + 1 =
313Impossible: if A=5, a= 5, P=1, p= 1, E=25, so e= 25.
John Strader: We found that you can make up the biggest number possible
GGEEAAPP = 312 because if there is a number that's a multiple of 5, there is
no P needed. You can make every multiple of 5 up to 310 using geap. You
can make a number ending in 3 or 4 by making number end in 5 and add a
certain number of little p's. For a number ending in 1 or 2, make a number
ending in 0 then add a certain number of big P's. The highest is 312 so the
next one you can't get is 313. You can't go bigger than 312 because you
are only allowed to use 2 of the same Geap letters. (example: Geapp is fine;
however Geappp is not fine).

Here's more (November
2000) work on the limit, this time from Maine

Note from the teacher: My students
noticed immediately that they were unable to make 313 in GEAP. The
difficult part was when I asked how they knew this was the smallest
number they couldn't make. They stuggled for several class periods
with this. Here is their final explanation. Karen Jagolinzer,
Harrison Middle School, Yarmouth, ME.
GEAP (From Mrs. Jagolinzer's 7th Grade Math Think Lab
Class)
G (g) = 125 E (e) = 25 A (a) = 5 P (p) = 1
Two numbers next to each other is addition. Capitalized
letters are positive numbers and lowercase letters are negative.
Therefore, "AA" is adding together two positive numbers, "aa" is
adding together two negative numbers, "Aa" is adding a negative
number to a positive number, and "aA" is adding a positive number to
a negative number. It is a given that "A" = 5. Since "PA" = 6, "P"
has to be 1. And also, if pA = 4, then "p" has to be 1. If "Ep" =
24, knowing that "p" = 1, "E" has to be 25. Since "PEG" = 151, and
"PE" = 26, "G" has to be 125.
We found that we could make the numbers 112 using just "P's"
and "A's". Then, if you add an "E", you can get any number between,
2512 and 25+12 using the "P's" and "A's" (positive and negative),
because "E" is 25 and if you can get all of the numbers 112 using
"A's" and "P's" than you can make 25 + (#'s 112) using just "A's",
"P's", and "E". By adding another "E" you can get the numbers 50 
(#'s 112) and 50 +(#'s 112). From there, you can add a "G" to get
"G"( #'s 162) and "G"+(#'s 162). This gives you all of the numbers
from 1 to 162. By adding another "G" you can get any number up to
312. If you put all of the numbers together (in positive form,
"GGEEAAPP") the highest number you can is 312 and if you can get all
of the numbers from 1312, but nothing higher, the lowest number you
can't make is 313.

Here is some work on the same issue by
a different group of students in the same class:

To find the answer we first found that the highest number you
can make on GEAP is GGEEAAPP (312), because you may only use two of
each letter. Then we found how to make the numbers 112. P=1, PP=2,
App=3, Ap=4, A=5, AP=6, APP=7, AApp=8, Aap=9, AA=10, AAP=11, AAPP=12,
After trying several strategies we came up with a chart that by using
the numbers 112 we could prove what the lowest number you couldn't
make on GEAP is.
We had already made 12 so basically itās 112 is the highest
you can go with just A and P so we added E (25). Then
E(25)aapp(12)=13 and if we can make 112 then we can make 1325 and
if we do uppercase instead of lowercase then that gives us the
numbers through 37, then we added the other E so itās EEaapp to get
us 38 then we do the same thing we did with the first E set bring us
up to 62(EEAAPP). That is the highest number we are able to make with
E, A, P so we then added the first G(125). We then repeated the
process like what we did with Eās and 12 only we did it with 62
giving us the numbers 1187. Finally we added the last G and we did
188(Ggeeaapp)312(GGEEAAPP).
312 GG+62
250 GG+62188 GG62
187 G+62
125 G
63 G62
62 EE+12
38 EE12
37 E+12
25 E
13 E12
So we are saying the lowest, positive, number you can't make
on GEAP is 313 because you can only make the numbers 1312.
Annie, Jon, and Adam

eeps comments: very interesting.
John
Strader's work, especially, is the
first one I've seen that begins to grapple successfully with WHY it's
possible to make all of the numbers up to some limit. The
Mainiacs from Yarmouth try to tackle it too. Which arguments
do you find to be the clearest? What would you have to do to make
them even more clear and convincing??

back to the Answer Book page

