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Solution to Puzzle 107.  BLOCKS

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 = p2

(2)   R = n2.  B+Y = m2.

(3)   R = r3.  B = b3.  Y is y2.

Combining (2) and (3) leads to

(4)   m2 = b3 +y2.

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 3-4-5 triangle.

32 + 42 = 52

Twice the above is

62 + 82 = 10or  36 + 64 = 100

Combining (1) and (2)

n2+m2 = p2

Since R =  n2= r3, 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 = n2.  This leads to B + Y = 36 = m2.

From (4), m2 = b3 +y2.  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

 

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The Universe in a Handkerchief: Lewis Carroll's Mathematical Recreations, Games, Puzzles, and Word Plays

Author:  Martin Gardner 

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