Stephen K. answered • 09/10/14
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Sondra,
The size of the squares doesn't matter, since the number of squares will be the same.
Think of the problem this way:
You have a row of 5 squares: 〈〈〈〈〈 (1) and on top of this you stack 5 more squares: 〈〈〈〈〈 (2)
and you do this for a total of 5 times so that you have 5 rows with 5 squares each:
〈〈〈〈〈 (5)
〈〈〈〈〈 (4)
〈〈〈〈〈 (3)
〈〈〈〈〈 (2)
〈〈〈〈〈 (1)
So how many squares do you have: You have 5 + 5 + 5 + 5 + 5 = 25 squares!
Stephen K.
Matt,
Given your interpretation of the problem, and I think you are correct, I only see one additional solution and that is the 1 large square to fill the entire 5 x 5 grid making a total of 26. Since 25 is only divisible by 5 and 1 I don't see any other combination that would "fill" the gird. Do you see another possible solution? Thanks for your input.
SK
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09/11/14
Stephen K.
I am stipulating here that all squares must be the same size, if we allow combinations of different size square, its still doable, just gets a bit messy.
SK
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09/11/14
Matt H.
Hi Stephen, thanks for your posts!
I'm interpreting the direction to mean that you can count any size square, whether it's 5 x 5, 4 x 4, etc., all the way out to the giant one.
If you allow for every size, AND you allow them to overlap, you'll get ...51 squares! :-)
I wish I could draw it out, but maybe try this: draw a 5 x 5 grid, and number the boxes 1-25, starting in the upper left and going left to right (so your second row starts with 6, etc.)
Now, notice you can make a 2 x 2 square out of boxes 1,2,6, and 7. Then you can overlap another one made out of 2,3,7, and 8.
If you follow this general idea and overlap at every possible place, you get a lot of boxes!
I came up with 51, but I may have miscounted.
Whatcha think?
THANKS,
Matt
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09/11/14
Stephen K.
You may be right, I played with it after I saw your note, didn't try to work overlapping squares, but made an assumption that all squares did not have to be the same size, in which case you get solutions such as 1-4x4 + 9-1x1, or 1-3x3 + 3-2x2+ 4-1x1,
etc. and I came up with 11 combinations of that sort. I'm not really sure what the "correct" approach actually is based on the way the problem is stated, it may well be a combination of "all-of-the-above".
Thanks for the input though, it was an interesting exercise.
Steve
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09/11/14
Avi B.
55 It follows the rule of the sum of 1^2+2^2+3^2.....
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01/20/20
Sunday B.
The answer is 55: 1x1 squares =25 2x2=16 3x3=9 4x4=4 5x5=1 25 + 16 + 9 +4 +1=55 total possible squares
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10/28/20
Anivia M.
The answer is 225
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03/12/21
Matt H.
09/11/14