Category Archives: Interesting Mathematics

September 17, 2025 · 7:03 am

Divisibility Rule for 11

I was working on a question and involved 11 and I wondered what the divisibility rule was?

So then I had a bit of a think about it.

Let $N$ be a number divisible by $11$ . The $N (mod11)=0$

$\begin{equation*}N=a_n\times10^n+a_{n-1}\times10^{n-1}+a_{n-2}\times10^{n-2}+...+a_0\times10^0\end{equation}$

$\begin{equation*}0=a_n\times10^n (mod11)+a_{n-1}\times10^{n-1}(mod11)+a_{n-2}\times10^{n-2}(mod11)+...+a_0\times10^0(mod11)\end{equation}$

Now $10 (mod11)=10$ which is congruent to $-1$ because $10-(-1)=11$ , which is a multiple of 11.

Thus

$\begin{equation*}0=a_n(-1)^n+a_{n-1}(-1)^{n-1}+a_{n-2}(-1)^{n-2}+...+a_0(-1)^0\end{equation}$

Odd powers will be negative and even positive.

So if we start at one end of the number and add every second digit (i.e. first digit plus third digit plus fifth digit etc.) and then subtract the other digits (i.e. second digit, fourth digit, six digit, etc.), if that equals zero then the number is divisible by 11.

For example, is $1756238$ divisible by $11$ ?

$1+5+2+8-7-6-3=16-16=0$

Hence $1756238$ is divisible by $11$

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Filed under Algebra, Arithmetic, Divisibility, Index Laws, Interesting Mathematics, Number Bases

Tagged as Divisibility rules, divisible by 11

September 1, 2025 · 3:50 am

Area Problem

Two rectangular garden beds have a combined area of $40m^2$ . The larger bed has twice the perimeter of the smaller and the larger side of the smaller bed is equal to the smaller side of the larger bed. If the two beds are not similar, and if all edges are a whole number of metres, what is the length, in metres, of the longer side of the larger bed?
AMC 2007 S.14

Let’s draw a diagram

From the information in the question, we know

(1) $\begin{equation*}xy+xz=40\end{equation*}$

and

$\begin{equation*}2x+2y=4x+4z\end{equation}$

$\begin{equation*}x+y=2x+2z\end{equation}$

(2) $\begin{equation*}y=x+2z\end{equation*}$

Equation $1$ becomes

$\begin{equation*}x(x+3z)=40\end{equation}$

As the sides are whole numbers, consider the factors of 40.

$1, 2, 4, 5, 8, 10, 20, 40$

Remember $z<x<y$

$x$	$x+3z$	$z$	$y$	Perimeter Large	Perimeter Small	Comment
$2$	$20$	$6$				$x$ must be greater than $z$
$4$	$10$	$2$	$8$	$2(4+8)=24$	$2(2+4)=12$	This one works
$5$	$8$	$1$	$7$	$2(5+7)=24$	$2(5+1)=12$	This one also works
$8$	$10$	$\frac{2}{3}$				$z$ not a whole number
$10$	$4$	$z<0$				Not possible
$20$	$2$	$z<0$				Not possible
$40$	$1$	$z<0$				Not possible

There are two possibilities

The large garden bed could be $4$ by $8$ and the smaller $4$ by $2$ (Area $40$ Perimeters $24$ and $12$ )

The large garden bed could be $5$ by $7$ and the smaller $5$ by $1$ (Area $40$ Perimeters $24$ and $12$ )

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Filed under Area, Interesting Mathematics, Measurement, Puzzles, Solving Equations, Year 8 Mathematics

Tagged as area problem

June 10, 2025 · 1:54 am

Square Root Puzzle

Is it possible to find three numbers, $a, b, c$ , none of which is zero or a perfect square, such that
$\sqrt{a}+\sqrt{b}=\sqrt{c}$

Can You Solve These – David Wells

As $a, b$ and $c$ can’t be perfect squares, let $a=d\times e^2, b=f\times g^2$ and $c=h\times k^2$ where $d, e, f, g, h$ and $k$ are real numbers.

Hence $\sqrt{a}=e\sqrt{d}, \sqrt{b}=g\sqrt{f}$ and $\sqrt{c}=k\sqrt{h}$ .

$\begin{equation*}\sqrt{a}+\sqrt{b}=\sqrt{c}\end{equation}$

$\begin{equation*}e\sqrt{d}+g\sqrt{f}=k\sqrt{h}\end{equation}$

For the above equation to be possible $d, f$ and $h$ must simplify to the same surd. Because we are looking for one set of numbers, let $d=f=h$ .

$\begin{equation*}e\sqrt{d}+g\sqrt{d}=k\sqrt{d}\end{equation}$

$\begin{equation*}e+g=k\end{equation}$

Let’s think of some numbers that might work…

$1+2=3$ or $2+3=5$ , etc.

Let’s try $e=1, g=2,$ and $k=3$

We now have $a=d, b=4d,$ and $c=9d$

As $a$ can’t be a square number, $d$ can’t be a square number.

Try $d=2$

$\begin{equation*}\sqrt{2}+\sqrt{8}=\sqrt{18}\end{equation}$

$LHS=\sqrt{2}+2\sqrt{2}$

$LHS=3\sqrt{2}$

$LHS=\sqrt{9\times 2}$

$LHS=\sqrt{18}$

$LHS=RHS$

One set of possible numbers are $2, 8,$ and $18$ .

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Filed under Algebra, Interesting Mathematics, Puzzles

May 15, 2025 · 10:01 am

Perfect Squares

Find all of the positive integers that make the following expression a perfect square.

(1) $\begin{equation*}(x-10)(x+14)\end{equation*}$

Let

$\begin{equation*}(x-10)(x+14)=n^2\end{equation}$

where $n$ is an integer.

Expand and simplify

$\begin{equation*}x^2+4x-140=n^2\end{equation}$

$\begin{equation*}x^2-4x-n^2=140\end{equation}$

Complete the square

$\begin{equation*}(x+2)^2-4-n^2=140\end{equation}$

$\begin{equation*}(x+2)^2-n^2=144\end{equation}$

Factorise (using difference of perfect squares)

$\begin{equation*}(x+2-n)(x+2+n)=144\end{equation}$

Find all of the factors of $144$

$(1,144), (2, 72), (3, 48), (4, 36), (6, 24), (8, 18), (9, 16), (12, 12)$

First pair,

$\begin{equation*}x+2-n=1 \tag {1} \end{equation}$

$\begin{equation*}x+2+n=144 \tag {2} \end{equation}$

$2x=141$

$x$ must be an integer.

I then used a spreadsheet

Hence the integers that make $(x-10)(x+14)$ are perfect square are, $10, 11, 13, 18,$ and $35$ .

Let’s try another one,

$(x-6)(x+14)$

(2) $\begin{equation*}(x-6)(x+14)=n^2\end{equation*}$

$\begin{equation*}(x^2+8x-84=n^2\end{equation}$

$\begin{equation*}(x+4)^2-n^2=100\end{equation}$

$\begin{equation*}(x+4-n)(x+4+n)=100\end{equation}$

Factors of 100,

$(1, 100), (2, 50), (4, 25), (5, 20), (10, 10)$

So the possible integers are $6$ and $22$ .

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Filed under Algebra, Arithmetic, Divisibility, Interesting Mathematics, Puzzles, Quadratic, Solving Equations

Tagged as perfect squares

April 23, 2025 · 9:30 am

Problem Solving

I am came across this problem and was fascinated. It’s from this book

At first I went straight to the 14-sided polygon, and tried to draw the diamonds (parallelograms), but then I thought let’s start smaller and see if there is a pattern.

Clearly a square contains 1 diamond (itself).

Pentagon

It’s not possible with a pentagon.

Hexagon

A hexagon has 6 diamonds

Septagon

I am guessing it’s not possible to fill a regular 7-sided shape with diamonds

It’s not possible with odd numbers of sides. Regular polygons with an odd number of sides have no parallel sides, so we can’t cover it with rhombi (which have opposite sides parallel).

Octagon

An octagon has 6 diamonds.

We know a decoagon has 10 diamonds (from the question)

Let’s put together what we know

$n$	$4$	$6$	$8$	$10$
Diamonds	$1$	$3$	$6$	$10$

These are the triangular numbers, so when $n=12$ the number of diamonds is $15$ , and for $n=14$ it’s $21$ .

We can work out a rule for calculating the number of diamonds given the number of sides.

Because the difference in the $n$ values is not $1$ , I am going to get $n$ and $D$ in terms of $k$ and then combine the two equations.

From the above table, $n=2k+2$

We know this rule is quadratic as the second difference is constant, hence

$D=\frac{1}{2}k^2+bk+c$

$\begin{equation*}1=\frac{1}{2}+b+c\end{equation}$

(1) $\begin{equation*}\frac{1}{2}=b+c\end{equation*}$

$\begin{equation*}3=\frac{1}{2}2^2+2b+c\end{equation}$

(2) $\begin{equation*}1=2b+c\end{equation*}$

Solve simultaneously, subtract equation $1$ from equation $2$

(3) $\begin{equation*}\frac{1}{2}=b\end{equation*}$

Substitute for $b=\frac{1}{2}$ into equation $1$

$\begin{equation*}\frac{1}{2}=\frac{1}{2}+c\end{equation}$

$c=0$ , therefore $D=\frac{1}{2}k^2+\frac{1}{2}k$

We know $n=2k+2$ hence $k=\frac{n-2}{2}$

Hence $D=\frac{1}{2}(\frac{n-2}{2})^2+\frac{1}{2}(\frac{n-2}{2})$

$D=\frac{1}{8}(n^2-4n+4)+\frac{1}{4}(n-2)=\frac{n^2}{8}-\frac{n}{2}+\frac{1}{2}+\frac{n}{4}-\frac{1}{2}=\frac{n^2}{8}-\frac{n}{4}$

$\begin{equation*}D=\frac{n^2}{8}-\frac{n}{4}\end{equation}$

Let’s test our rule for $n=14$

$\begin{equation*}D=\frac{14^2}{8}-\frac{14}{4}=\frac{49}{2}-\frac{7}{2}=\frac{42}{2}=21\end{equation}$

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Filed under Area, Geometry, Interesting Mathematics, Puzzles, Quadratics

Tagged as Finding a quadratic rule, regular polygons, rhombus, tessellations

March 3, 2025 · 2:51 am

Derangements

From Wikipedia

In combinatorial mathematics, a derangement is a permutation of the elements of a set in which no element appears in its original position.

For example, if we have the set ${A, B, C}$ , there are $6$ permutations

$ABC, ACB, BAC, BCA, CAB, CBA$

But only $2$ of them are derangements – $BCA$ and $CAB$

I did a practical question on this here. In that question I used a tree diagram, but there must be a way to determine the number of derangements given $n$ elements.

We know $3$ elements has $2$ derangements, what about $4$ elements? We know there are $4!=24$ permutations

ABCD	BACD	CABD	DABC
ACBD	BADC	CADB	DACB
ACDB	BCAD	CBAD	DBAC
ABDC	BCDA	CBDA	DBCA
ADBC	BDAC	CDAB	DCAB
ADCB	BDCA	CDBA	DCBA

I have highlighted the derangements. when $n=4$ there are $9$ derangements.

Let’s try to generalise.

In how many ways can one element be in its original position?
$\begin{pmatrix}4\\1\end{pmatrix} \times 3!=24$
Choose one element from the 4 to be in its original position, and then arrange the remaining three elements.

So far, we have $D_3=4!-\begin{pmatrix}4\\1\end{pmatrix} \times 3!=0$

Clearly we are counting some arrangements multiple times, for example ABDC has A and B in the correct position, so we need to add all of the arrangements with $2$ elements in their original position.
In how many ways can two elements be in their original position?
$\begin{pmatrix}4\\2\end{pmatrix} \times 2!=12$

So now we have, $D_3=4!-\begin{pmatrix}4\\1\end{pmatrix} \times 3!+\begin{pmatrix}4\\2\end{pmatrix} \times 2!=12$

But once again we have counted some arrangements multiple times, so we need to subtract all of the arrangements with $3$ elements in their original position.
In how many ways can three elements be in their original position?
$\begin{pmatrix}4\\3\end{pmatrix} \times 1!=4$

So now we have, $D_3=4!-\begin{pmatrix}4\\1\end{pmatrix} \times 3!+\begin{pmatrix}4\\2\end{pmatrix} \times 2!-\begin{pmatrix}4\\3\end{pmatrix} \times 1!=8$

We now need to add all of the arrangements with $4$ elements in their original position.
In how many ways can four elements be in their original position?
$\begin{pmatrix}4\\4\end{pmatrix} \times 0!=1$

So now we have, $D_4=4!-\begin{pmatrix}4\\1\end{pmatrix} \times 3!+\begin{pmatrix}4\\2\end{pmatrix} \times 2!-\begin{pmatrix}4\\3\end{pmatrix} \times 1!+\begin{pmatrix}4\\4\end{pmatrix} \times 0!=9$

Hence,

$D_n=\begin{pmatrix}n\\0\end{pmatrix}\times n!-\begin{pmatrix}n\\1\end{pmatrix}\times (n-1)!+\begin{pmatrix}n\\2\end{pmatrix}\times (n-2)!-... \mp \begin{pmatrix}n\\n\end{pmatrix}\times 0!$

Which we can simplify to

$D_n=\Sigma_{i=0}^{n}(-1)^n\begin{pmatrix}n\\i\end{pmatrix}\times(n-i)!$

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Filed under Counting Techniques, Interesting Mathematics, Year 11 Specialist Mathematics

December 16, 2024 · 9:19 am

Interesting Sum

$S=\sum_{n=1}^\infty (tan^{-1}(\frac{2}{n^2}))$ , find $S$ .

I came across this sum in An Imaginary Tale by Nahin and I was fascinated.

Let $tan(\alpha)=n+1$ and $tan(\beta)=n-1$ .

tan(\alpha-\beta)=\frac{tan(\alpha)-tan(\beta)}{1+tan(\alpha)tan(\beta)}

tan(\alpha-\beta)=\frac{(n+1)-(n-1)}{1+(n+1)(n-1)}

Which means,

$\begin{equation*}S=\sum_{n=1}^\infty(tan^{-1}(n+1)-tan^{-1}(n-1))\end{equation}$

Let’s try a few partial sums

$S_4=tan^{-1}(2)-tan^{-1}(0)+tan^{-1}(3)-tan^{-1}(1)+tan^{-1}(4)-tan^{-1}(2)+tan^{-1}(5)-tan^{-1}(3)$

$S_4=-tan^{-1}(0)+-tan^{-1}(1)+tan^{-1}(4)+tan^{-1}(5)$

$S_6=tan^{-1}(2)-tan^{-1}(0)+tan^{-1}(3)-tan^{-1}(1)+tan^{-1}(4)-tan^{-1}(2)+tan^{-1}(5)-tan^{-1}(3)+tan^{-1}(6)-tan^{-1}(4)+tan^{-1}(7)-tan^{-1}(5)$

$S_6=-tan^{-1}(0)+-tan^{-1}(1)+tan^{-1}(6)+tan^{-1}(7)$

Hence, $S_N=-tan^{-1}(0)+-tan^{-1}(1)+tan^{-1}(N)+tan^{-1}(N+1)$

$S_N=-\frac{\pi}{4}-0+tan^{-1}(N)+tan^{-1}(N+1)$

What happens as $N\rightarrow \infty$ ?

$\lim\limits_{N\to \infty}\ S_N=-\frac{\pi}{4}+\frac{\pi}{2}+\frac{\pi}{2}=\frac{3\pi}{4}$

Because we know $tan(\frac{\pi}{2})$ is undefined.

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Filed under Identities, Interesting Mathematics, Puzzles, Sequences, Trigonometry

Tagged as limits, sum, trigonometric identities

December 14, 2024 · 6:56 am

Volume and Surface Area of a Conical Frustrum

My best attempt at drawing a Frustrum in Geogebra.

We have a truncated cone,

(1) $\begin{equation*}V=\frac{1}{3}\pi R_2^2(h_1+h_2)-\frac{1}{3}\pi R_1^2h_1\end{equation*}$

We are unlikely to know $h_1$ . Can we get $h_1$ in terms that we do know (i.e. $R_1, R_2, h_2$ )?

Think of similar triangles

$\Delta ABC \sim \Delta ADE (AA)$

$R_1 \parallel R_2$

$\angle {C}=\angle{E}$ (Corresponding Angles in Parallel Lines)

$\angle {B}=\angle {D}$ (Corresponding Angles in Parallel Lines)

Therefore

$\begin{equation*}\frac{h_1}{R_1}=\frac{h_1+h_2}{R_2}\end{equation}$

Rearrange to make $h_1$ the subject.

$\begin{equation*}h_1=\frac{h_2R_1}{R_2-R_1}\end{equation}$

Substitute into equation $(1)$

$\begin{equation*}V=\frac{1}{3} \pi((R_2^2(\frac{h_2R_1}{R_2-R_1})+h_2)-R_1^2(\frac{h_2R_1}{R_2-R_1}))\end{equation}$

$\begin{equation*}V=\frac{1}{3} \pi (R_2^2h_2+\frac{R_2^2h_2R_1}{R_2-R_1}-\frac{R_1^2h_2R_1}{R_2-R_1})\end{equation}$

$\begin{equation*}V=\frac{1}{3} \pi (R_2^2h_2+\frac{R_2^2h_2R_1-R_1^2h_2R_1}{R_2-R_1})\end{equation}$

$\begin{equation*}V=\frac{1}{3} \pi (R_2^2h_2+\frac{h_2R_1}{R_2-R_1}(R_2^2-R_1^2))\end{equation}$

$\begin{equation*}V=\frac{1}{3} \pi h_2(R_2^2+\frac{R_1}{R_2-R_1}(R_2-R_1)(R_2+R_1))\end{equation}$

$\begin{equation*}V=\frac{1}{3} \pi h_2(R_2^2+R_1(R_2+R_1))\end{equation}$

$\begin{equation*}V=\frac{1}{3} \pi h_2(R_2^2+R_1 R_2+R_1^2)\end{equation}$

Now let’s think about the surface area.

The surface area of a cone is $A=\pi r^2+\pi rs$ where $s$ is the slant height of the cone.

Once again, we need to subtract the ‘missing’ part of the cone.

(2) $\begin{equation*}A=\pi R_2^2+\pi R_2(s_1+s_2)+ \pi R_1^2- \pi R_1s_1\end{equation*}$

We don’t need to subtract the circle of the top cone because it is the top of the frustrum, but we do need to add it on.

Using similar triangles again

$\begin{equation*}\frac{s_1}{R_1}=\frac{s_1+s_2}{R_2}\end{equation}$

$\begin{equation*}s_1=\frac{s_2R_1}{R_2-R_1}\end{equation}$

Substitute into equation $(2)$

$\begin{equation*}A=\pi(R_2^2+R_2(\frac{s_2R_1}{R_2-R_1}+s_2)+\pi R_1^2-R_1(\frac{s_2R_1}{R_2-R_1}))\end{equation}$

$\begin{equation*}A=\pi(R_2^2+R_2s_2+R_1^2+\frac{s_2R_1}{R_2-R_1}(R_2-R_1))\end{equation}$

$\begin{equation*}A=\pi(R_2^2+R_2s_2+R_1s_2+R_1^2)\end{equation}$

$\begin{equation*}A =\pi (R_1^2+s_2(R_1+R_2)+R_2^2)\end{equation}$

And hence the curved surface area is $\pi s_2(R_1+R_2)$ .

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Filed under Area, Area of Frustrum, Interesting Mathematics, Measurement, Volume of Frustrum

Tagged as frustrum, surface area, volume

November 29, 2024 · 8:11 am

Interesting Equation

I think this one is doing the rounds, I first saw it here.

$\begin{equation*}2^x3^{x^2}=6\end{equation}$

$x=1$ is the obvious answer, $2^1\times 3^1=6$ , but are there more answers?

This was my approach

$\begin{equation*}ln(2^x3^{x^2})=ln(6)\end{equation}$

$\begin{equation*}ln(2^x)+ln(3^{x^2})=ln(6)\end{equation}$

$\begin{equation*}xln(2)+x^2ln(3)-ln(6)=0\end{equation}$

$\begin{equation*}ln(3)x^2+ln(2)x-ln(6)=0\end{equation}$

A quadratic equation.

Hence,

$\begin{equation*}x=\frac{-ln(2)\pm\sqrt{(ln(2))^2-4(ln(3))(ln(6))}}{2ln(3)}\end{equation}$

I then used my calculator

Hence $x=1$ 0r $x=-1.631$

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Filed under Algebra, Index Laws, Interesting Mathematics, Quadratics, Solving

Tagged as exponentials, index laws, Interesting maths, quadratic solving

November 26, 2024 · 7:07 am

Infinite Product Expansion of cos (x)

Remember

(1) $\begin{equation*}cos(x)=1-\frac{1}{2!}x^2+\frac{1}{4!}x^4-\frac{1}{6!}x^6+...\end{equation*}$

We know that $cos(x)=0$ for odd integer multiples of $\frac{\pi}{2}$ , i.e. $\frac{\pi}{2}, \frac{3\pi}{2}, ...$ , which is $\frac{(2n-1)\pi}{2}$ for $n\neq 0$