Prove that for any whole number N > 5, a square can be divided into N squares.
(I suspect this is the type of problem where you either see the answer fairly quickly or it takes you hours to see the solution. I was clearly in the last category.)
Squares

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Re: Squares
Ok... for even N's, take a square from the lower right corner and adjust its size so that the remaining space can be filled with an odd number of squares. The larger the corner square, the more small squares can fill the rest.DanishDynamite wrote:Prove that for any whole number N > 5, a square can be divided into N squares.
(I suspect this is the type of problem where you either see the answer fairly quickly or it takes you hours to see the solution. I was clearly in the last category.)
For odd N's... cut out a large square in the center of the bottom edge. Adjust size to need as with the even solution.
Hardly rigorous, but I think it makes it clear enough.

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Re: Squares
"Hardly rigorous" seems right. :Dgnome wrote:Ok... for even N's, take a square from the lower right corner and adjust its size so that the remaining space can be filled with an odd number of squares. The larger the corner square, the more small squares can fill the rest.DanishDynamite wrote:Prove that for any whole number N > 5, a square can be divided into N squares.
(I suspect this is the type of problem where you either see the answer fairly quickly or it takes you hours to see the solution. I was clearly in the last category.)
For odd N's... cut out a large square in the center of the bottom edge. Adjust size to need as with the even solution.
Hardly rigorous, but I think it makes it clear enough.
I can understand your solution for even N's as it is the one I found, but your solution for odd N's leaves me puzzled.
Could you post a diagram?

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Re: Squares
...DanishDynamite wrote:"Hardly rigorous" seems right. :Dgnome wrote:Ok... for even N's, take a square from the lower right corner and adjust its size so that the remaining space can be filled with an odd number of squares. The larger the corner square, the more small squares can fill the rest.
For odd N's... cut out a large square in the center of the bottom edge. Adjust size to need as with the even solution.
Hardly rigorous, but I think it makes it clear enough.
I can understand your solution for even N's as it is the one I found, but your solution for odd N's leaves me puzzled.
Could you post a diagram?
Actually I'm incorrect, you still get an even number. Let me work on it some more.

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Trying to hide this in a spoiler ('select' it, or CtrlA to view)
<table bgcolor=white><tr><td>
Given any existing arrangement, any existing square can be subdivided into four equal ones, yielding an extra three squares in total. So if we can solve for any three consecutive numbers, then all numbers greater than that are proven soluble by induction.
<pre>++++<br>   <br>++++<br>   = 6 (and therefore 9, 12, 15...)<br> ++<br>  <br>+++
++++<br>   <br>+++ <br>   <br>++++ = 7 (and therefore 10, 13, 16...)<br>  <br>  <br>  <br>+++
+++++<br>    <br>+++++<br>  <br> ++ = 8 (and therefore 11, 14, 17...)<br>  <br> ++<br>  <br>+++<br></pre></td></tr></table>
<table bgcolor=white><tr><td>
Given any existing arrangement, any existing square can be subdivided into four equal ones, yielding an extra three squares in total. So if we can solve for any three consecutive numbers, then all numbers greater than that are proven soluble by induction.
<pre>++++<br>   <br>++++<br>   = 6 (and therefore 9, 12, 15...)<br> ++<br>  <br>+++
++++<br>   <br>+++ <br>   <br>++++ = 7 (and therefore 10, 13, 16...)<br>  <br>  <br>  <br>+++
+++++<br>    <br>+++++<br>  <br> ++ = 8 (and therefore 11, 14, 17...)<br>  <br> ++<br>  <br>+++<br></pre></td></tr></table>

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