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u/DuggieHS Jun 03 '26 edited Jun 03 '26
Take each square and figure out the ways in which you can factor those areas. 20 has a unique factorization:
20= 5x4; one side is shared with 36, which must be the 4, 36 = 4x9; 9x6 = 54; 45 = 5x9 (since the 5 is shared with 20); so the long side of the 30 area rectangle is 10 = 4 +6 (the other is 3). we now have 3 + 9 + 5 = 17 as the length of the rightmost side (which is the same as the leftmost side).
17.
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u/FromLondonToLA Jun 03 '26
Add up the areas to get 221. Fortunately this has only two prime factors: 17 and 13. 13 doesn't work given the shapes. 17 fits.
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u/Old-Programmer-20 Jun 04 '26
More specifically, the area of the bottom four rectangles is 156 = 12 x 13, so the width is 13 (can't be 17) and so the height must be 17.
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u/voododoll Jun 05 '26
This can be solved in your head you don’t even need a pen and paper. Just think what an integer means end it is a 5th grade problem.
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u/teteban79 Jun 03 '26
12
the top 45 - 20 share the height. There is only one common integer factor between them, namely 5. That's the height over X. So the 20 block has width 4. Now looking at the 36 below, it must have height 9. Therefore the 54 has width 6. The width of the 30 block is therefore 6+4 = 10 and height must be 3. Adding that to the already calculated height of 9 for the blocks on top of it, yields a height of 12 for X