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

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?


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Do the squares have to have integer sides? Because nine squares, each of side 5/3, is a perfectly fine solution.

This 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.

And that's what we got to through trial and error. Just wanted to check I wasn't missing anything obvious.

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

But personally I think the first answer is correct; there is no requirement for integer sides in the question.

I've printed it off to entertain my children.

I'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...

I agree.

My 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 not specify 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.

Having said which, there is a 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.

Your solution is the one I saw immediately, but I then concluded that nine even squares drives home the *don't put constraints in the question that aren't there* more soundly. Neither of my children have solved this yet.

I'm not sure if nine even squares, or a scaled version of the previous option, best underlines the issue. But yes. I think the solution relies on first guessing and then proving that there is no integral solution, before looking to break the rule that wasn't actually specified.

An interesting little problem, and rather devious.


Oh 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

One four by four square, 9 1x1 squares in the rest of the spaces.

Sod it - it's 9 total, isn't it. Damn!

I am sorry to say that the authors of the book have admitted a typo - should be 8 squares. We all had eight squares easily!

This makes more sense because otherwise you just copy the example in the question scaled down.

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