Oops, then it's simply,
(1^2)+(2^2)+(3^2)+...+(8^2)
A nice easy, sort of, chess puzzle.
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vincent_pang wrote:
Oops, then it's simply,
(1^2)+(2^2)+(3^2)+...+(8^2)
...which adds up to 204.
vincent_pang wrote:
Oops, then it's simply,
(1^2)+(2^2)+(3^2)+...+(8^2)
That's the number of squares, what about all possible rectangles (squares of course being rectangles)
When you said squares, I assumed rectangles with all sides of equal length. For that is how it is said.
1(1+2+3+4+5+6+7+8)+2(2+3+4+5+6+7+8)+3(3+4+5+6+7+8)+...+8(8)
750? A rather aesthetic number.
vincent_pang wrote:
When you said squares, I assumed rectangles with all sides of equal length. For that is how it is said.
Yes, but we finished that like a year ago. Recently, we've been working on the number of overall rectangles (see post #16 by nimzovich)
vincent_pang wrote:
750? A rather aesthetic number.
You're going to have to defend your answer. My answer (see posts #17 and #20) was 1296, so at least one of us is definitely wrong.
Ok, how did you come up with that number?
I could try to explain my answer, but I wouldn'd know where to start.
vincent_pang wrote:
Ok, how did you come up with that number?
I could try to explain my answer, but I wouldn'd know where to start.
dmeng wrote:
You're going to have to defend your answer. My answer (see posts #17 and #20) was 1296, so at least one of us is definitely wrong.
You're going to have to defend your answer. My answer (see posts #17 and #20) was 1296, so at least one of us is definitely wrong.
vincent_pang wrote:
I could try to explain my answer, but I wouldn't know where to start.
That's definitely a problem.
@ orangehonda - Thanks.
tyzebug wrote:
No doubt the proof is too large for the margin to contain.
That made my day. 
A far cry from a proper proof, however:
# of rectangles of height 1 = 8 x # of rectangles of height 1 in an 8x1 matrix
# of rectangles of height 2 = 7 x # of rectangles of height 2 in an 8x2 matrix
# of rectangles of height 3 = 6 x # of rectangles of height 3 in an 8x3 matrix
...
# of rectangles of height 8 = 6 x # of rectangles of height 3 in an 8x8 matrix
(the multiplier in each of the above being the number of possible starting positions along the y-axis for a rectangle of that height)
# of rectangles of height 1 in an 8x1 matrix
= # of rectangles of height 2 in an 8x2 matrix
= # of rectangles of height 3 in an 8x3 matrix
...
= # of rectangles of height 8 in an 8x8 matrix
= 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 (the number of possible starting positions along the x-axis for a rectangle of each width starting with 1)
= 36
So, we get:
8 x 36
+7 x 36
+6 x 36
...
+1 x 36
= 36 x (8 + 7 + 6 + 5 + 4 + 3 + 2 + 1)
= 36x36
= 1296
Tut tut. It's 1968.
The formula is (x^2)[(9-x)^2)], 0<9 where x is an integer and is the length of the sides of the square.
There are many lengths including;
1x1, 2x2, 3x3, ... , 8x8
You must add all the possibilities.
(1^2)[(9-1)^2)]+(2^2)[(9-2)^2)]+(3^2)[(9-3)^2)]+...+(8^2)[(9-8)^2)]
There,
1968 if I'm not mistaken.
Your formula finds the sum of the areas of the squares, not the number of the squares themselves.