My colleague's 12 year was set question 5. The external square does not count and everything must be a square. We tried overlapping and about 20 different permutations but could not get 9 squares. I said I'd post it here because there are smart people on LJ.

Any thoughts?

#### A hard maths problem my colleague's 12 year old was set

May 21st, 2014

bohemiancoastThis isn't doable with integer square sides, as follows:

You cannot have a square of size 4x4, because that leaves 9, which only divides 4/4/1, 4/1/1/1/1/1 or 1/1/1/1/1/1/1/1/1 for solutions with 4, 7 and 10 smaller squares respectively. Each time you replace a 2x 2 square with 4 1x1 squares, you increase the number of squares by 3.

You can only have 1 square of size 3x3 for fit reasons. That leaves 16, which can be split into 4, 7, 10, 13 or 16 squares, for solutions with 5,8,11,14 or 17 squares in total.

So the largest square is 2x2, and by the same approach as before that will generate solutions with 25, 22, 19, 16, 13, 10 or 7 squares. None of those have 9 squares.

So there is no 9-square split with integer sides.

attimes_bracingWill post the answer when my friend's kid gets it from their teacher.

bohemiancoastbohemiancoastI'm reminded of the time that Marianne was given, aged about 8, the homework 'how many different ways can you make 79p up out of smaller coins'. I *did* solve it, but it took several pages of explanation and most of the weekend...

bellinghmanMy quick bit of algebra also indicates that you can't do it with integral values:

x * 1 + y * 4 + z * 9 -> 25

x + y + z -> 9

z <= 1 because you can't fit two 3x3 squares in without an overlap. And I'm not bothering with any 4x4 squares, since simple inspection means the minimum total number of squares is then 10

if z = 1, then

x + 4y -> 16

x + y -> 8

And therefore 3y -> 8 - not an integral value.

Contrarily, if z = 0

x + 4y -> 25

x + y -> 9

And 3y -> 16, still not an integral solution.

So my solution is to abandon integer values, since the problem does

notspecify them, and give the same design as the original one shown (just each 'unit' is 5/6 as large). You can rotate the bottom right quadrant or exchange it with one of the others as desired.bellinghmanisa sneaky solution that involves a 3x3 in one corner, 3 2x2s and 4 1x1s to fill up the rest of the area, and then drawing a 1x1 in the centre of the 3x3. All edges are now on the grid.But that's a cheat in my opinion.

bohemiancoastbellinghmanAn interesting little problem, and rather devious.

bellinghmanOh well, that means

If z = 0

x + 4y -> 25

x + y -> 8

And 3y -> 15, so y -> 5

Can't fit 5 2x2s in, so that fails

If z = 1, then

x + 4y -> 16

x + y -> 7

And therefore 3y -> 9. Promising. y -> 3

Yep, One 3x3, 3 2x2s beside it, and you end up with 4 1x1 squares. Lots of permutations, none difficult to make

ffuturesffuturesattimes_bracingms_cataclysm