Dividing Squares

Some of the university lecturers here were interested in the question:

Is it possible to divide a given square into n squares for any n≥6?

(Obviously all squares neednt be the same size)

16 Answers

62
Lokesh Verma ·

"here" ?

1
fibonacci ·

yes i think because let area of larger square=n2 sq units
it can be divided into n smaller squares of area
1 sq unit + 3 sq units + 5 sq units + ... + 2n-1 sq units
such that their sum is = n2

1
fibonacci ·

example for n=7, n2=49
it can be divided into 7 squares of area
1 sq unit + 3sq units + 5 sq unit + ... + 13 sq units = 49 sq units

sir am i correct?

341
Hari Shankar ·

"here" meant where i stay :D

@fibonacci - no i meant like u can divide the square into 6 squares or 7 squares or any number of squares greater than 6.

341
Hari Shankar ·

Well? Heck this is not so tough. You need minimal maths background to be able to do this.

39
Dr.House ·

is it right way to proceed??

we could subdivide the square in 4, 9, 16 or ... squares, then subdivide them further etc.

this way you get every n with n = 1 mod 3, every n with n = 1 mod 8, every n with n = 1 mod 15 etc.

you can get every n such that n-1 is a sum of 3's, 8's, 15's, etc
this solves all n≥ 17

n=16 is clear

62
Lokesh Verma ·

bhargav.. even though I did not understand your explanation "fully", I think what you are tryign to say (or what i am interpreting seems to suggest that you are correct. )

1
Philip Calvert ·

how do we explain n = 6 this way ?
also maybe n = 8 ?

[2] maybe i didnt get your method.

or is there a misprint in the question ?

341
Hari Shankar ·

there is no misprint. please dont let this thread die!

1
palani ............... ·

@bhargav

wat do you mean by 1 mod 3

11
Devil ·

Basically we need to confirm for cases 8,11 and 14.
Is that right prophet sir?

341
Hari Shankar ·

I wouldnt know unless you explain your line of thought to me.

39
Dr.House ·

nothing spoken of my approach here??

no comments prophet sir?

341
Hari Shankar ·

ya, first how to we go from the statement: "we can sub-divide into 4,9,16 squares ..." to "this way you get for all n = 1 mod 3 ..."?

next: from this point to "sub-division for every n≥17 is can be obtained this way"?

11
Devil ·


So, now if we keep on dividing the bigger square in each of the figures A,B,c in 4 parts, 3 additional squares get added during each subsequent division......So, that gives we can divde a square in n squares for all n≥6.

341
Hari Shankar ·

Yeah that's correct.

We first note that if you have a sub-division into n squares, you also have a sub-division into n+3 squares by choosing any one square and dividing it into 4 parts.

So if you find a division for n = 6,7,8 you would have proved for all n

n = 7 is easy because you can divide into n = 4 and so you can divide into 4+3 = 7

for n = 6, consider a 3X3 and then consider a 2X2 in it as one square. That gives you 9 -4+1 = 6 squares

for n = 8, consider a 4X4 and consider a 3X3 as one square. Now you have 16-9+1 = 8 squares and so we have proved for all n

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