Download The Grid Rectangle Problem

Math 485
Summer 2006
The Grid Rectangle Problem
 Choose one number of tiles and explain how you know you have all of
the grid rectangles for that number.
 Share one of your patterns with your partner. Describe the pattern and
try to explain why it occurs.
How did we divide up the work? Why isn’t the group display in consecutive order?
To think about: What would be an ideal way to arrange the collective display to meet
our mathematical goals?
How do you know you have all of the grid rectangles for a particular
number?:
To come up with how many you can get, you can figure out the number of factors.
E.g., 35 can be divided by 1, 5, 7, 35 and there are only four ways to express it: 1 x 35,
5 x 7, 7 x 5, 35 x1.
How to know when you’ve found all of the factors? Start with 1 (Is the number divisible
by 1?); 2 (is the number divisible by 2?). If so, you list the other number. When you
get to one you’ve already written, you are done.
Example: 24
1 x 24
2 x 12
3x8
4x6
5: can’t do 5
6: already have 6, so have all of the factors
Figure out all of the numbers used in that list, listing them in pairs: 1, 24, 2, 12, 3, 8, 4,
6
Counted all those numbers, told you how many ways you could express the number of
rectangles. 8 factors of 24, so 8 grid rectangles.
Chelsea’s Claim: To find the number of factors, you start with 1 (is the number divisible
by 1?), if so then list the factor pair. Then check 2, 3, etc. Continue that until you get
to a factor pair that you’ve already listed.
Two factor partners make a factor pair.
Claim: The number of factors determines the number of rectangles.
With 24, you know 5 can’t be a factor because you can’t repeatedly add 5 to get to 24.
24 is not a multiple of 5. 5 doesn’t divide evenly into 24.
Math 485
Summer 2006
Could choose some numbers and test Chelsea’s theory…
School math terms vs. math as a discipline terms vs. invented terms
basic fact
factor
factor partner
algorithm is (loosely) a rule or set of steps that you go through to produce a desired
result (we will come back to this term…)
Testing Chelsea’s claim:
Report on 28:
1: 1x28
2: 2x14
3: no because 3 doesn’t divide into 28 without a remainder
4: 4x7
5: no because 28 is not divisible by 5
6: no because 28 is not divisible by 6
7: repeat the third solution, by swapping pairs (4x7=7x4). so there are no other
solutions
As soon as you get to the second factor being smaller than the first factor in the pair
(e.g., 7 x 4 – 4 is smaller than 7) so you’ve already exhausted all of your factors.
Things to think about: Can you test Chelsea’s claim all the way out? Why does it work?
Is there another way to find all of the factors of a number? Is there a method that is
maximally efficient (i.e., doesn’t use extra steps)?
Some patterns in the grid rectangle problem:
The area of the rectangle equals the number of tiles you have to work with. (e.g., 24:
the area of a 4x6 rectangle is 24). All the rectangles for that number have the same
area.
If you take the list of factors:
1
20
4
5
2
10
You have 3 sets there, multiply that by 2 to get the number of grid rectangles.
Unless it’s a perfect square, you multiply by 2 and then subtract 1.
9:
Math 485
Summer 2006
1
3
9
3
Multiply that by 2 to get 4, and subtract 1.
36:
1
2
3
4
6
36
18
12
9
6
Claim/conjecture: Multiply by the number of factor pairs by 2 (2 x 5= 10). But because
one of them is a square you need to subtract 1. (4 x9 and 9x4 are two distinct
rectangles, but if you flip 6x6 you get the same rectangle).
Claim/conjecture: For any given number of tiles, if that number is a perfect square,
then the number of rectangles it will generate will be an odd number of rectangles. All
other target numbers that are not perfect squares will have an even number of
rectangles.
Notice that we haven’t proven these claims yet…
Claim/conjecture: Prime numbers will have only two rectangles. (e.g., 11 has 1 x 11
and 11 x 1; 7 has 1 x 7 and 7 x1)
A prime number only has one and itself as factors. A prime number has exactly two
factors.
What about 1? Is 1 prime? It has 1 and itself, but 1 is itself.
If prime numbers have two rectangles, 1 would be aberrant because it doesn’t have 2
rectangles.
You can break any number down into prime numbers: 36 = 2x2x3x3 Might the prime
factorization help us with this problem?
Definition of prime number: A prime number has exactly two different factors: 1 and
itself.
If definition said: A prime number has a maximum of two different factors: 1 and itself,
then 1 wouldn’t be ruled out.
Mathematical convention: 1 is not prime.
Math 485
Summer 2006
Robin’s claim/conjecture: You have the number (which is the product), the range of
factors is between 1 and the product. For every number you will have either have 2, 4,
or 6 factors plus the number of perfect squares within that range.
For example: 100:
the 6 factors: 1, 2, 5, 10, 20, 50
plus the 3 perfect squares: 4, 25, 100
so that’s 9 total factors.
Post-class note: Robin wants to revise her conjecture. She noticed that the 2, 4, and 6
don’t work for every number. Maybe it has something to do with whether the product
is perfect square…