Puzzles
Wool circles
\(n\) people stand in a circle. The first person takes a ball of
wool, holds the end and passes the ball to his right, missing a
people. Each person who receives the wool holds it and passes the
ball on to their right, missing \(a\) people. Once the ball returns to
the first person, a different coloured ball of wool is given to
someone who isn't holding anything and the process is repeated. This is
done until everyone is holding wool.
For example, if \(n=10\) and \(a=3\):
In this example, two different coloured balls of wool are needed.
In terms of \(n\) and \(a\), how many different coloured balls of
wool
are
needed?
Show answer & extension
Hide answer & extension
Starting with the person who starts with the wool and
going anti-clockwise, number the people \(0,1,2,3,4,...\). As the
wool is passed, it will be held by people with numbers:
$$0,a+1,2(a+1),3(a+1),...,k(a+1),...$$
The first person will have the wool again when
$$k(a+1)\equiv 0 \mathrm{\ \ mod\ } n$$
or
$$k(a+1)=ln.$$
This will first occur when (hcf is highest common factor):
$$l=\frac{a+1}{\mathrm{hcf}(a+1,n)}\mathrm{\ \ and\ \ }k=\frac{n}{\mathrm{hcf}(a+1,n)}$$
\(k\) is also the number of people who are holding the wool. So
the number of different coloured balls needed is:
$$\frac{n}{\left(\frac{n}{\mathrm{hcf}(a+1,n)}\right)}$$
$$=\mathrm{hcf}(a+1,n)$$
Extension
The ball is passed around the circle of \(n\) people again. This
time,
the number of people missed alternates between \(a\) and \(b\). How many
different coloured balls of wool are now needed?
Sum equals product
\(3\) and \(1.5\) are a special pair of numbers, as \(3+1.5=4.5\)
and
\(3\times 1.5=4.5\) so \(3+1.5=3\times 1.5\).
Given a number \(a\), can you find a number \(b\) such that
\(a+b=a\times b\)?
Show answer & extension
Hide answer & extension
If \(a+b=a\times b\), then:
$$ab-b=a$$
$$b(a-1)=a$$
$$b=\frac{a}{a-1}$$
This will work for any \(a\not=1\) (\(a=1\) will not work as this
will
mean
division by zero).
Extension
(i) Given a number \(a\), can you find a number \(b\) such that
\(b-a=\frac{b}{a}\)?
(ii) Given a number \(a\), can you find a number \(b\) such that
\(b-a=\frac{a}{b}\)?
(iii) Given a number \(a\), can you find a number \(b\) such that
\(a-b=\frac{b}{a}\)?
(iv) Given a number \(a\), can you find a number \(b\) such that
\(a-b=\frac{a}{b}\)?
Bézier curve
1) A set of points \(P_0\), ..., \(P_n\) are chosen (in the example \(n=4\)).
2) A set of points \(Q_0\), ..., \(Q_{n-1}\) are defined by \(Q_i=t P_{i+1}+(1-t) P_i\) (shown in green).
3) A set of points \(R_0\), ..., \(R_{n-2}\) are defined by \(R_i=t Q_{i+1}+(1-t) Q_i\) (shown in blue).
.
.
.
\(n\)) After repeating the process \(n\) times, there will be one point. The Bézier curve is the path traced by this point at \(t\) varies between 0 and 1.
What is the Cartesian equation of the curve formed when:
$$P_0=\left(0,1\right)$$
$$P_1=\left(0,0\right)$$
$$P_2=\left(1,0\right)$$
Show answer & extension
Hide answer & extension
If
$$P_0=\left(0,1\right)$$
$$P_1=\left(0,0\right)$$
$$P_2=\left(1,0\right)$$
then
$$Q_0=t\left(0,0\right)+(1-t)\left(0,1\right)=\left(0,1-t\right)$$
$$Q_1=t\left(1,0\right)+(1-t)\left(0,0\right)=\left(t,0\right)$$
and so
$$R_0=t\left(t,0\right)+(1-t)\left(0,1-t\right)=\left(t^2,(1-t)^2\right)$$
which means that
$$x=t^2$$$$y=(1-t)^2$$
or
$$y=(1-\sqrt{x})^2$$
Extension
What should \(P_0\), \(P_1\) and \(P_2\) be to get a curve with Cartesian equation
$$y=(1-\sqrt{2x})^2$$
Folding A4 paper
A Piece of
A4 paper is folded as shown:
What shape is made?
Show answer & extension
Hide answer & extension
The sides of A4 paper are in the ratio \(1:\sqrt{2}\). Let the width of the paper be 1 unit. This means that the height of the paper is \(\sqrt{2}\) units.
Therefore on the diagram, \(AE=1, BE=1, AC=\sqrt{2}\). By Pythagoras' Theorem, \(AB=\sqrt{2}\), so \(AB=AC\).
\(BF=DF=\sqrt{2}-1\) so by Pythagoras' Theorem again, \(BD=2-\sqrt{2}\). \(CD=1-(\sqrt{2}-1)=2-\sqrt{2}\). Hence, \(CD=BD\) and so the shape is a kite.
Extension
Prove that for a starting rectangle with the sides in any ratio, the resulting shape is a
cyclic quadrilateral.
Multiples of three
If the digits of a number add up to a multiple of three, then the number is a multiple of three. Therefore if a two digit number, \(AB\) (first digit \(A\), second digit \(B\); not \(A\times B\)), is a multiple of three, then \(A0B\) is also a multiple of three.
If \(AB\div 3=n\), then what is \(A0B\div 3\)?
Dartboard
Concentric circles with radii 1, \(\frac{1}{2}\), \(\frac{1}{3}\), \(\frac{1}{4}\), ... are drawn. Alternate donut-shaped regions are shaded.
What is the total shaded area?
Show answer & extension
Hide answer & extension
The shaded area is:
$$\pi (1)^2 - \pi (\frac{1}{2})^2 + \pi (\frac{1}{3})^2 - \pi (\frac{1}{4})^2 + \pi (\frac{1}{5})^2 - ...$$
$$=\sum_{i=1}^\infty \frac{\pi (-1)^{i-1}}{i^2}$$
$$=\pi\sum_{i=1}^\infty \frac{(-1)^{i-1}}{i^2}$$
$$=\pi\left(\frac{\pi^2}{12}\right)$$
$$=\frac{\pi^3}{12}$$
Extension
Prove that
$$=\sum_{i=1}^\infty \frac{(-1)^{i-1}}{i^2}=\frac{\pi^2}{12}$$
Seven digits
"I'm thinking of a number. I've squared it. I've squared the square. And I've multiplied the second square by the original number. So I now have a number of seven digits whose final digit is a 7," said Dr. Dingo to his daughter.
Can you work out Dr. Dingo's number?
Show answer & extension
Hide answer & extension
Let's call Dr. Dingo's number \(n\). If the number is squared twice then multiplied by \(n\), we get \(n^5\).
For all integers \(n\), the final digit of \(n^5\) is the same as the final digit of \(n\). In other words:
$$n^5\equiv n \mod 10$$
Therefore, the final digit of Dr. Dingo's number is 7.
$$7^5=16807$$
$$17^5=1419857$$
$$27^5=14348907$$
So, in order for the answer to have seven digits, Dr. Dingo's number was 17.
Extension
For which integers \(m\) does there exist an integer \(n\) such that for all integers \(x\):
$$x^n\equiv x \mod m$$
Parabola
On a graph of \(y=x^2\), two lines are drawn at \(x=a\) and \(x=-b\) (for \(a,b>0\). The points where these lines intersect the parabola are connected.
What is the y-coordinate of the point where this line intersects the y-axis?
Show answer & extension
Hide answer & extension
The co-ordinates of the points where the lines intersect the parabola are \((a,a^2)\) and \((-b,b^2)\). Hence the gradient of the line between them is:
$$\frac{a^2-b^2}{a-(-b)}=\frac{(a+b)(a-b)}{a+b}=a-b$$
Therefore the y-coordinate is:
$$b^2 + b(a-b) = ba$$
Ferdinand Möbius, who discovered this property called the curve a Multiplicationsmaschine or 'multiplication machine' as it could be used to perform multiplication.
Extension
How could you use the graph of \(y=x^2\) to divide 100 by 7?
![](javascript:showlimage("woolcircles.png"))
![](javascript:showlimage("bezier.gif"))
![](javascript:showlimage("folding-a4-paper.png"))
![](javascript:showlimage("kite.png"))
![](javascript:showlimage("dartboard.png"))
![](javascript:showlimage("parabola.png"))