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A quaint incongruent integral partitioning conundrum (Posted on 2006-12-20) Difficulty: 3 of 5
What is the maximum number of rectangles into which a 17x24-rectangle can be partitioned, when all rectangles must be incongruent and have integer side lengths?

  Submitted by JLo    
Rating: 4.0000 (1 votes)
Solution: (Hide)
The maximum is 35. When adding up all integers that represent areas of integer-sided rectangles, going from smaller to larger and counting integers multiple times, depending on the number of differently shaped rectangles they represent, one gets the following sum:

1+2+3+4+4+5+6+6+7+8+8+9+9+10+10+11+12+12+12+
13+14+14+15+15+16+16+16+17+18+18+18+19+20+20+20 =
= 408 = 17*24

So no more than 35 different rectangles can be made out of a 17x24-rectangle. To verify that this is indeed possible, one has to solve a not-so-difficult jigsaw puzzle. A possible solution is this:

==========oo===++-=+-+=-
==========oo---++-=+-+=-
---------+oo---++-=+-+=-
---------+oo---++-=+-+=-
ooo=ooooo-oo---++-=+-+=-
ooo=----==oo===++-=+-+=-
---=oooo==oo===++-=+-+=-
---=oooo--++===++-=+-+=-
---=oooo--++oooooo=+-+=-
---=oooo--++oooooo=+-+=-
---+++++--++-------+-+=-
---+++++--=========+-+=-
===+++++ooooooooooo+-+=-
===+++++============-+=-
===oooooooooooooooooo+=-
===-------------------=-
===++++++++++++++++++++-

Comments: ( You must be logged in to post comments.)
  Subject Author Date
No SubjectRosepark2023-10-02 09:44:21
Puzzle AnswerK Sengupta2023-10-02 09:29:48
Solutionre: CeilingEric2006-12-20 15:54:43
CeilingEric2006-12-20 15:34:14
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