MHB Solving a problem using a simplified version

  • Thread starter Thread starter estex198
  • Start date Start date
AI Thread Summary
Carmelo is tasked with creating a decorative wall for the X Games using a 21x21 tile array forming a large X pattern. The calculation for the number of colored tiles reveals that 79 tiles are needed, derived from counting the colored tiles in the top and bottom rows, and the diagonals. A formula for any odd n x n grid is proposed as C(n) = 4n - 5, which accounts for the filled rows and intersecting diagonals. The discussion highlights the inadequacy of using cross multiplication for this problem, prompting inquiries about more formal solving methods. Understanding the pattern and logic behind the arrangement is crucial for accurate calculations.
estex198
Messages
14
Reaction score
0
While I know the answer to this problem, I can't figure out *how* to get it.

Carmelo has been commissioned to create a decorative wall for the 21st Annual X Games consisting of a square array of square tiles in a pattern forming a large X. The following example shows a pattern with 5 rows and 5 columns. If the wall will have a similar pattern with 21 rows and 21 columns, how many of the colored tiles will be needed?

View attachment 4368

With the 5x5 array there are 25 total blocks and 15 colored blocks. I thought maybe I could use ratios and cross multiply: 15/25 = x/441, but this isn't giving me 79. I know 79 is the correct answer not only because its shown in the back of the book, but because I actually took a ruler and completed the pattern on paper (tedious as it was, I was really frustrated and needed to know I wasnt wasting my time.) Please help!
 

Attachments

  • problem1.png
    problem1.png
    5 KB · Views: 99
Physics news on Phys.org
We know we will have the top and bottom rows colored for a total of 42 tiles. Then one diagonal will be 19 (21 minus the two already counted) and the other diagonal will be 18 for a total of 42 + 19 + 18 = 79. :D

Using this same logic, we could derive a formula for an $n\times n$ grid (where $3\le n$ and $n$ is odd). If we let $C(n)$ denote the number of colored blocks for such a grid, we would get:

$$C(n)=n+n+(n-2)+(n-3)=4n-5$$
 
Rusty: So your logic is that since two rows will be filled in (n + n), the diagonals will equal the remaining number of lines minus two, with the exception that one of the lines will be one less (n-2) + (n-3), since it will intersect with the other line. Thanks a million btw for a clear and direct answer!
I'm still curious however, if there is a more formal method of solving this problem? Can anyone explain why cross multiplying ratios will not work? Thanks in advance!
 
I was reading documentation about the soundness and completeness of logic formal systems. Consider the following $$\vdash_S \phi$$ where ##S## is the proof-system making part the formal system and ##\phi## is a wff (well formed formula) of the formal language. Note the blank on left of the turnstile symbol ##\vdash_S##, as far as I can tell it actually represents the empty set. So what does it mean ? I guess it actually means ##\phi## is a theorem of the formal system, i.e. there is a...
Back
Top