Question 604355: A rectangular floor that measures 78 inches by 102 inches
is to be completely covered with square tiles. The tiles are
available in sizes with only whole-number side lengths. What
is the smallest number of uncut tiles that could be used to
cover the floor?
Answer by KMST(5328)   (Show Source):
You can put this solution on YOUR website! The width of the floor can be covered by 1-inch wide tiles, using 78 tiles.
We can use 2-inch wide tiles, using 39 of them, because (2 inch) x 39 = 78 inch.
We could also use 3-inch wide tiles (26 of them) because 3 x 26 = 78.
We could even use 6-inch wide tiles (13 of them) because 6 x 13= 78.
In fact we could cover the width of the floor with tiles of any number of inches in width, as long as that number is a factor of 78 inches. Even 13, or 26, or 39 inches would work.
However, we want the width of the tiles (in inches) to also be a factor of 102, the length of the room in inches
102 is divisible by 1, 2, 3, and 6 (102 = 6 x 17), but it is not evenly divided by 13, 26, or 39.
The greatest common factor of 78 and 102 is 6.
With 6 inch by 6 inch tiles, we would fit 17 tiles along the 102 inch length of the room and 13 tiles along the 78 inch width.
So we would have 13 rows of 17 tiles (or 17 rows of 13 tiles, depending on how you look at it).
In any case, we would have 13 times 17 tiles, for a total of 221 tiles, with the least number of the biggest square tiles we can get.
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