EPTIn story telling mode:
I give you a square, and says "Hey man, can you cut it into 14 squares for me?" You should be able to do it. And I can say "now do 15, now do 18, ..." You should be able to do that as well.
Again, put answer in spoilers
--evan
[Math Puzzle] 21
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evanthebouncy!
United States12796 Posts
In story telling mode: I give you a square, and says "Hey man, can you cut it into 14 squares for me?" You should be able to do it. And I can say "now do 15, now do 18, ..." You should be able to do that as well. Again, put answer in spoilers --evan | ||
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Chairman Ray
United States11903 Posts
+ Show Spoiler + Even numbers, cut one large square from the whole, and then split the remaining part into squares. The thinner the remaining part, the more squares can be cut. This will support any number of squares of an even number excluding 2. For odd numbers, the large square that you initially cut, split it into 4 squares and do the same thing. This will work for any odd number excluding 3 and 5. | ||
Empyrean
17035 Posts
http://oeis.org/A074764 Does this count ?EDIT: I think Chairman Ray's explanation is correct. | ||
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nanoscorp
United States1237 Posts
+ Show Spoiler + Turning one square into 4 is easy: Just split it down and across the middle. So, given any size square we can make it into 3 more. This process is repeatable as well. Now, we need to figure out how to make 0, 1 and 2 (mod 3) squares. First, 0mod3: Think Tic-Tac-Toe. Making 9 squares initially with 4 cuts means we can make 12, 15, 18, etc. Next, 1: Again, we go back to the simplest split, down and across the middle. From 4 initial squares we can make 7, 10, 13... For 2mod3: Make a 4x4 grid and take a 3x3 large square in one corner. You're left with 7 1x1 squares, meaning 8 total. From 8, we can do 11, 14, 17... | ||
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ReketSomething
United States6012 Posts
On January 30 2012 15:27 nanoscorp wrote: I'll try: + Show Spoiler + Turning one square into 4 is easy: Just split it down and across the middle. So, given any size square we can make it into 3 more. This process is repeatable as well. Now, we need to figure out how to make 0, 1 and 2 (mod 3) squares. First, 0mod3: Think Tic-Tac-Toe. Making 9 squares initially with 4 cuts means we can make 12, 15, 18, etc. Next, 1: Again, we go back to the simplest split, down and across the middle. From 4 initial squares we can make 7, 10, 13... For 2mod3: Make a 4x4 grid and take a 3x3 large square in one corner. You're left with 7 1x1 squares, meaning 8 total. From 8, we can do 11, 14, 17... wow, pro. | ||
thedeadhaji
39489 Posts
On January 30 2012 15:27 nanoscorp wrote: I'll try: + Show Spoiler + Turning one square into 4 is easy: Just split it down and across the middle. So, given any size square we can make it into 3 more. This process is repeatable as well. Now, we need to figure out how to make 0, 1 and 2 (mod 3) squares. First, 0mod3: Think Tic-Tac-Toe. Making 9 squares initially with 4 cuts means we can make 12, 15, 18, etc. Next, 1: Again, we go back to the simplest split, down and across the middle. From 4 initial squares we can make 7, 10, 13... For 2mod3: Make a 4x4 grid and take a 3x3 large square in one corner. You're left with 7 1x1 squares, meaning 8 total. From 8, we can do 11, 14, 17... ah damn, couldn't figure out the last part! | ||
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stenole
Norway869 Posts
What we know. A) Any square can be cut into 4 squares by cutting horizontally and vertically. B) Any square can be cut into 6 squares by cutting one 2/3*2/3 square and five 1/3*1/3 squares. By applying A to an arbitrary square, we will end up with 3 more squares than we had before. Same with B except we end up with 5 more squares. We start off with 1 square initially. By applying B 2 times and A x times, we can get any number in the sequence 11, 14, 17, 20... By applying B 1 time and A x times, we can get any number in the sequence 12, 15, 18, 21... By applying B 0 times and A x times, we can get any number in the sequence 10, 13, 16, 19... | ||
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Klockan3
Sweden2866 Posts
Meh, you can add 2 squares to any 4 square lattice by changing it into 6 squares, a 6 square by changing it into an 8 square etc. You get 6 squares by making a 9 square lattice and then removing the cross in a 4 square, continue adding 2 squares in the same way by extending the edges while keeping one large square. So thus you can easily add 2 squares to anything originally containing a 4 square lattice, so you just need to find an even and an odd containing a 4 square lattice. We already got the trivial 4 square lattice, and if we take the 4 square and make a 4 square in one of the corners we get 7 squares. Done. | ||
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xxpack09
United States2160 Posts
On January 30 2012 15:44 thedeadhaji wrote: ah damn, couldn't figure out the last part! Same :[ That's pro stuff | ||
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Glacierz
United States1245 Posts
On January 30 2012 15:00 Chairman Ray wrote: I think I got an answer, not sure if it's correct + Show Spoiler + Even numbers, cut one large square from the whole, and then split the remaining part into squares. The thinner the remaining part, the more squares can be cut. This will support any number of squares of an even number excluding 2. For odd numbers, the large square that you initially cut, split it into 4 squares and do the same thing. This will work for any odd number excluding 3 and 5. This is by far the simplest approach. + Show Spoiler + Let N = number of squares required, and the initial square having a length of 1. The length of the first square to cut can be calculated by: 1-2/N if N is even 1-2/(N-3) if N is odd (which is then to be divided into 4 equal pieces) Then just cut along the the remaining L shape for squares until done. | ||
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fritfrat
United States50 Posts
I just did it empirically.. the same way as stenole, except a different way of getting to 12, since I came up with the L shaped one last. Cutting a square into 4 pieces adds 3 squares, cutting a square into 9 pieces adds 8 squares, and cutting a square into one big one and 5 little ones adds 5 squares. You get 10 by 1+3+3+3, you get 11 by 1+5+5, and you get 12 by 1+3+8. Add multiples of 3 to any number above that and you're set. This doesn't count for the last possible number, which is 8- for 8 you need another "corner-border representation" of 1 big square and a L of 7 little squares, but it answered the question I agree, Chairman's approach is the simplest/best | ||
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