1. David quickly arranged all of these blocks
to form a square. 2. With the all
red blocks, he formed a larger square array. With the rest of the blocks,
he formed a smaller square array.
3. With all the red
blocks, he formed a cube. With all the blue blocks, he formed another
cube. With all the yellow blocks, he formed a square array.
What is the smallest number of red blocks the
set can have? What is the smallest number of blue blocks the set can
have? What is the smallest number of yellow blocks the set can have?
Solution:
Let R be the number of red cubes, B be the number of blue cubes,
and Y be the number of yellow cubes. Then
(1) R+B+Y = p^{2}
(2) R = n^{2}. B+Y = m^{2}.
(3) R = r^{3}. B = b^{3}.
Y is y^{2}.
Combining (2) and (3) leads to
(4) m^{2} = b^{3} +y^{2}.
The solution can be done by trying different perfect
square and perfect cube numbers on the equations above. A clue can be
inferred from the knowledge of the following property of the 345 triangle.
3^{2} + 4^{2} = 5^{2}
Twice the above is
6^{2} + 8^{2} = 10^{2 }or
36 + 64 = 100 Combining
(1) and (2) n^{2}+m^{2}
= p^{2} Since R =
n^{2}= r^{3}, one
number that is both a cube and a square is 64, since 64 is square of 8 and cube
of 4. So try R = 64 = n^{2}. This leads to B + Y = 36 = m^{2}.
From (4), m^{2} = b^{3} +y^{2}.
Since 36 = 27 + 9, it can be inferred that B is 27 and Y = 9.
Thus, there are 64 red cubes, 27 blue cubes and 9 yellow
cubes
