SOLUTION: A rectangular room is completely tiled by 1-foot square tiles. All the adjacent to a door or wall are purple, and the rest of the tiles are white. If exactly 2/7 of the tiles are p

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Question 1204445: A rectangular room is completely tiled by 1-foot square tiles. All the adjacent to a door or wall are purple, and the rest of the tiles are white. If exactly 2/7 of the tiles are purple, then what is the smallest possible area of the room, in square feet?
Found 3 solutions by Edwin McCravy, greenestamps, mccravyedwin:
Answer by Edwin McCravy(20064)   (Show Source): You can put this solution on YOUR website!
Suppose the total number of 1 ft2 squares is A = LW
We don't want to count the 4 corner squares twice, so since the length is L,
Then there are 2(L-2) non-corner squares along the 2 lengths and 2(W-2) 
non-corner squares along the 2 widths. So there are 2(L-2)+2(W-2) + 4 corner
squares.  That's 2L-4+2W-4+4 = 2L+2W-4 squares. 







Complete the rectangle.  We want to find what we must add
to both sides so that LW-7-7W will factor in the form

(L+p)(W+q)
Multiply that out:
LW + qL + pW + pq
We see q=-7, p=-7, pq=49
Sow have to add 49 to both sides

 


  

The factors of 35 are

35x1 = 7x5 Two possibilities if L > W

L-7=35, W-7=1 or L=42, W=8, LW = 336
L-7=7, W-7=5 or L=14, W=12, LW = 168

The smallest possible area is 168 ft2.

Edwin
Answer by greenestamps(13209)   (Show Source): You can put this solution on YOUR website!


Let L and W be the length and width of the room.

Then the total number of tiles is LW.

And the number of purple tiles is 2L+2W-4. The -4 is because 2L+2W together counts the 4 corner squares twice.

The purple tiles are 2/7 of the total:





This is a Diophantine equation -- two variables with only one equation; but the number of solutions is finite because the variables are non-negative integers.

Here is a solution to the problem using standard methods for solving linear Diophantine equations.

Solve the equation for one variable in terms of the other.





Make it easier to find integer solutions by performing the division on the right and expressing the result as quotient and remainder.



In that form of the equation, 7 is an integer, and L has to be an integer; that means has to be an integer. And that means (W-7) is a factor of 35.

The factors of 35 are 1, 5, 7, and 35. We are to find the minimum value of the product LW.

To do that, we find the product LW for each of the possible values of (W-7).
  W-7  W   L   LW
 -----------------
   1   8  42  336
   5  12  14  168
   7  14  12  168
  35  42   8  336

Note in the table that the dimensions and area of the room in the second half of the table repeat those in the first half. That is because the original equation is symmetric in the two variables L and W. If we note that early in our work, then we know we don't have to complete the whole table.

ANSWER: The smallest possible area of the room in square feet is 168.
Answer by mccravyedwin(409)   (Show Source): You can put this solution on YOUR website!
I kept making a mistake completing the rectangle, analogous to
completing the square. What do we have to add to both sides of 
LW-7L-7W = -14 to make the left side factor as (L+?)(W+?)? I
finally found my error and now the above solution is complete
and correct in case you saw my solution before I corrected it.

Edwin McCravy
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