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We received a number of good solutions to this problem. Most of you noticed that the key to making rectangles that work for all bases is factorising quadratic equations. Luke from London Oratory School gave a good explanation of this:
The green section is a square, so its area is equal to $x^2$.

The red section consists of two rectangles with dimensions $a, x$ and $b, x$. Therefore the

red area is equal to $ax+bx$, which equals $x(a+b)$.

We can see that the area of the blue section will always have dimensions $a$ and $b$, so its area is equal to $ab$, if it ʻﬁllsʼ the gap created by the red area.

The total area is equal to the sum of these component areas.

Thus you can make a rectangle for all bases for equations of the form $x^2 + x(a+b) + ab$ where $a$ and $b$ are positive integers. The rectangle has dimensions $(x+a)$ by $(x+b)$.

So for the rectangle $x^2 + 7x + 12$, we can see that 1) $7=a+b$ and 2) $12=ab$.

$a=3 , b=4$ satisﬁes these equations.

So for any base we can make the rectangle with dimensions $(x+3)$ by $(x+4)$.

Vanessa and Annie sent us this solution:

Using 1 square and 12 sticks: You can make 7 different rectangles which work in all bases. These are:

$x(x+12)$, $(x+1)(x+11)$, $(x+2)(x+10)$, $(x+3)(x+9)$, $(x+4)(x+8)$, $(x+5)(x+7)$, $(x+6)(x+6)$.

You can make rectangles for all bases with dimensions $(x+A)(x+B)$ where $A$ and $B$ are positive integers and $A+B=12$.

Using the same logic, if you have one square and 100 sticks, you can make 51 rectangles that work for all bases:

$x(x+100)$, $(x+1)(x+99)$, $(x+2)(x+98)$, ..., $(x+50)(x+50)$.

Using 1 square and 12 units: You can make 3 different rectangles which work in all bases:

$(x+1)(x+12)$, $(x+2)(x+6)$, $(x+3)(x+4)$.

You can make rectangles for all bases with dimensions $(x+A)(x+B)$ where $A$ and $B$ are positive integers and $AB=12$.

Using the same logic, if you have one square and 100 sticks, you can make 5 rectangles which work in all bases: $(x+1)(x+100)$, $(x+2)(x+50)$, $(x+4)(x+25)$, $(x+5)(x+20)$, $(x+10)(x+10)$.

Using one square, $p$ sticks, $q$ units, you can only make a rectangle which works for all bases if $p=A+B$ and $q=AB$ for $A$ and $B$ positive integers.

Extension: Using $n$ squares, we can make rectangles which work in all bases of dimensions $(ax+b)(cx+d)$ when $a$,$b$,$c$,$d$ are positive integers and $ac=n$. Then we will use $ad+bc$ sticks and $bd$ units. This will give rectangles which have the squares arranged in a rectangle. But we could arrange the squares in different ways, like L-shapes.

Well done!

The red section consists of two rectangles with dimensions $a, x$ and $b, x$. Therefore the

red area is equal to $ax+bx$, which equals $x(a+b)$.

We can see that the area of the blue section will always have dimensions $a$ and $b$, so its area is equal to $ab$, if it ʻﬁllsʼ the gap created by the red area.

The total area is equal to the sum of these component areas.

Thus you can make a rectangle for all bases for equations of the form $x^2 + x(a+b) + ab$ where $a$ and $b$ are positive integers. The rectangle has dimensions $(x+a)$ by $(x+b)$.

So for the rectangle $x^2 + 7x + 12$, we can see that 1) $7=a+b$ and 2) $12=ab$.

$a=3 , b=4$ satisﬁes these equations.

So for any base we can make the rectangle with dimensions $(x+3)$ by $(x+4)$.

Vanessa and Annie sent us this solution:

Using 1 square and 12 sticks: You can make 7 different rectangles which work in all bases. These are:

$x(x+12)$, $(x+1)(x+11)$, $(x+2)(x+10)$, $(x+3)(x+9)$, $(x+4)(x+8)$, $(x+5)(x+7)$, $(x+6)(x+6)$.

You can make rectangles for all bases with dimensions $(x+A)(x+B)$ where $A$ and $B$ are positive integers and $A+B=12$.

Using the same logic, if you have one square and 100 sticks, you can make 51 rectangles that work for all bases:

$x(x+100)$, $(x+1)(x+99)$, $(x+2)(x+98)$, ..., $(x+50)(x+50)$.

Using 1 square and 12 units: You can make 3 different rectangles which work in all bases:

$(x+1)(x+12)$, $(x+2)(x+6)$, $(x+3)(x+4)$.

You can make rectangles for all bases with dimensions $(x+A)(x+B)$ where $A$ and $B$ are positive integers and $AB=12$.

Using the same logic, if you have one square and 100 sticks, you can make 5 rectangles which work in all bases: $(x+1)(x+100)$, $(x+2)(x+50)$, $(x+4)(x+25)$, $(x+5)(x+20)$, $(x+10)(x+10)$.

Using one square, $p$ sticks, $q$ units, you can only make a rectangle which works for all bases if $p=A+B$ and $q=AB$ for $A$ and $B$ positive integers.

Extension: Using $n$ squares, we can make rectangles which work in all bases of dimensions $(ax+b)(cx+d)$ when $a$,$b$,$c$,$d$ are positive integers and $ac=n$. Then we will use $ad+bc$ sticks and $bd$ units. This will give rectangles which have the squares arranged in a rectangle. But we could arrange the squares in different ways, like L-shapes.

Well done!