Using 12 squares make as many different rectangles as possible. what is the anwser.
Also what is the answer if you use 15 squares, 11 squares, 24 squares, 9 squares, 7 squares
Providing that you must use all twelve squares in each rectangle, and you must show the entire surface of each square, you can try the brute force method.
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For 12 squares:
1 x 12
2 x 6
3 x 4
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1 x 7 is unique
1 x 11 is unique
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The others have several possibilities, as in the 12 squares problem.
If you have a better plan, I would like to see it.
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The challenge now is to find an algorithm by which you can determine the number of rectangles possible given the number of squares, but without actually having to list each possibility.
Even more interesting would be writing a simple proof that such an algorithm is true, again without listing all the possible configurations for each problem.
For example, how many rectangles can be made from 4456 squares?
I would not want to go one by one on this, and it almost certainly is amenable to a reasonable algorithmic solution without the aid of software.
Please let me know if you have an answer.
Thank you.
Bye.
I would have to guess the simplest scenario, that all the squares are exactly alike. I think if you are allowed to use squares of any size, not all alike, the number of solutions is potentially infinite.
For example, suppose you have one large square and eleven small squares that fit together in a line exTactly as long as a side of the large square.
The possiblities are infinite and largely a waste of time.
2 x 6 = 6 x 2, so with squares that are all alike both of these arrangements are counted as the same thing, i.e., it is impossible to distinguish order under the given constraints.
It is now time to take my daughter down the street to dance class.
Bye for now.

Hello, hjohnson7968!
Using 12 squares, make as many different rectangles as possible. What is the answer?
What if you use 15 squares, 11 squares, 24 squares, 9 squares, 7 squares?
The question is: how many divisors does 12 have?
12 is divisible by 1, 2, 3, 4, 6, and 12.
Assuming arectangle is distinct from a
, six rectangles are possible.
There is a formula for this situation.
To find the number of divisors of an integer,
. . write its prime factorization: .
The number of divisors is: .
(Add one to each exponent and multiply)