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## Question

I have a chess board that has 13 squares on a side.

I want to break it [along the borders of the individual squares] into smaller square pieces.

I could make a 12x12 square and 25 1x1 squares - 26 in all.

I could make a 7x7 square, 3 6x6 squares and 12 1x1 squares - 16 in all.

What is the fewest I could make?

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One 13x13 square.

"Want" doesn't equal "have to".

Okay, lame answer. I have to go to work. Will work on this there.

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I can do it with 13. One 8x8, two 5x5, three 3x3, and 7 2x2. I don't know if this is the ultimate answer, but I suspect the final solution would have zero 1x1 squares.

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12

8x8, 3- 5x5's, 2 - 3x3's, 2- 2x2's, 4- 1x1's

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12

8x8, 3- 5x5's, 2 - 3x3's, 2- 2x2's, 4- 1x1's

12 is the smallest number, although I did it with 1 7x7, 2 6x6, 1 5x5, 5 2x2 and 3 1x1 squares.

I've quickly come to learn that the next question will be: what is the optimal solution for a board of dimensions nxn, where n is odd.

(n+11)/2, for n > 3. The division consists of 1 ((n+1)/2)x((n+1)/2) square, 2 ((n-1)/2)x((n-1)/2) squares, 1 ((n-3)/2)x((n-3)/2) square, (n-3)/2 2x2 squares, and 3 1x1 squares.

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Eleven.

1 - 7x7

2 - 6x6

1 - 4x4

1 - 3x3

3 - 2x2

2 - 1x1

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I only count 160 sq. units of area there, rather than 169.

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Eleven.

1 - 7x7

2 - 6x6

1 - 4x4

2 - 3x3

3 - 2x2

2 - 1x1

I only count 160 sq. units of area there, rather than 169.

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I got twelve. 2- 7x7s, 4-2x2, 2-4x4s and 4-3x3s

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I got twelve. 2- 7x7s, 4-2x2, 2-4x4s and 4-3x3s

Since any 7x7 division includes the center square, you can have only one of them.

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WELL...

If you cut it 0 times you get 1 square thats 13 x 13,

but something tells me thats not the answer you were looking for...

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WELL...

If you cut it 0 times you get 1 square thats 13 x 13,

but something tells me thats not the answer you were looking for...

Pieces is plural.

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