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Two Cubes

Two cubes, each with integral side lengths, have a combined volume equal to the total of the lengths of their edges. How big are the cubes? [If you find a result by 'trial and error' you'll need to prove you have found all possible solutions.]

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Common Divisor

Find the largest integer which divides every member of the following sequence: 1^5-1, 2^5-2, 3^5-3, ... n^5-n.

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Novemberish

a) A four digit number (in base 10) aabb is a perfect square. Discuss ways of systematically finding this number. (b) Prove that 11^{10}-1 is divisible by 100.

Factorising with Multilink

Stage: 4 Challenge Level: Challenge Level:1
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!