Archive for December, 2014

Tests of divisibility

December 11, 2014 Leave a comment

Today I was marking my year 9 classes books and came across some work on prime factor decomposition and tests of divisibility. Yesterday I had been arguing with Colin Beveridge (@icecolbeveridge) about the use of formula triangles, (see the comment section on this post) and any other method of anything that was presented in the fashion “do it this way and don’t worry about the why.” I felt a wave of hypocrisy flow over me. I had never explained the tests of divisibility to the class, and furthermore I didn’t even know myself why they worked! I thought I’d explore them and see what I came up with.

If the last digit is divisible by 2, the number is divisible by 2.

This test of divisibility is obvious, the definition of an even number is that it has 2 as a prime factor, and all even numbers have a last number divisible by 2. No further exploration needed. I know why this one works and I’m certain the class do too.

If the sum if the digits is a multiple of 3, then the number is divisible by 3.

This is a fact I’ve known since I was at primary school back in the 1980s, but I can honestly say that I’ve never thought about why it works. I wrote a couple if equations out, realised I should be working in modulo 3 and came up with this:


Which basically boils down to the fact any number can be split into the digits multiplied by a powerful of ten.

A= (x0)10^0 + (x1)10^1 +… + (xn)10^n

As (10^n) mod 3 = 1 for all natural numbers (and 0) then it follows that:

(A)mod 3 = (x0 +x1+…+xn) mod 3

Which implies our test of divisibility. This also implies the test if divisibility for 9 (ie is the sum of digits is divisible by 9 then so is the number) as (10^n)mod9 = 1 for all natural numbers and 0. To prove you just follow the exact working but in mod 9.

A number is divisible by 4 if it’s last two digits are divisible by 4.

This one uses the fact that if two numbers are both divisible by a number then so is their sum.


We know any number bigger than 100 can be expressed 100a + b where a and b are natural numberso and b<100. We also know that as 4 is a factor of 100, 100a is divisible by 4 if a is a natural number. Hence the whole number is divisible by 4 iff the best is divisible by 4, and b is the last two digits.

From this we can deduce the test of divisibility for 8 (ie a number larger than 1000 is divisible by 8 iff the last three digits are divisible by 8) as 1000 is divisible by 8. The proof is the same, but you split the number into 1000a + b where a and b are natural numbers and b is less than 1000.

Testing for 5 and 6

Testing for 5 (the last number is 5 or 0) is another obvious one that needs no further exploration. And the test for 6 is simple, if 2 and 3 are prime factors 6 must be a factor, as all numbers can be expressed as a product of their prime factors and 6 is the product of 2 and 3.

Testing divisibility for 10

Again, this is obvious as we are talking about a base ten system! That leaves just one more test.

To test for divisibility by 7, take the last digit, double it. Take this away from the number you are left with if you subtract the last number and divide by ten. If the answer is 0 or divisible by 7 (ie if the answer mod 7 = 0) then the original number us divisible by 7.

What a ridiculously complicated divisibility test. It looks much nicer algebraically:

(10a + b) mod 7 = 0 (a and b are natural numbers, b < 10)


(a -2b)mod7 = 0

Once I expressed it algebraically I had two simultaneous equations and solved like this:


I enjoyed working through them and now I’m confident I know why these tests work, but I’m fairly sure that the ones that aren’t obvious would be too complicated to explain to year 9. This made me think about their use, and I certainly think it’s fine to use them even without understanding. Which brought me back to the formula triangles debate. This is one that will rage, but as I explained in my post on it, I don’t have a great issue with people who understand the algebra using them. I also don’t have a great issue with people who struggle at maths and aren’t going to pursue it past GCSE using them. My issue is with higher attainers who could and should understand the algebra being taught then with no deeper conceptual understanding. I guess it’s a topic I will need to think more on.

Area of a semi-circle – the return

December 8, 2014 3 comments

Last week I wrote this piece in which I relayed my thought process when solving this problem from Ed Southall (@solvemymaths).


After sharing it a lot of people tweeted me their solutions and I thought I’d look at a few here. I think the variety of solutions people used, and the amount of people who engaged with this puzzle is testament both to how interesting a puzzle this is and also how fantastic maths is. If you did it a different way, do let me know!

My solution was fairly long compared to many of the others. I think this maybe down to my heavy preference to Trigonometry, and my brain’s insistence on using it if possible! Here are some other, more concise methods:

Using the Tangent Ratio

In my solution I used right angled triangle trigonometry to find the lengths of the chords, then used Pythagoras’s Theorem to find the diameter. Steve Atkinson (@Small_Ears) also used trig, spotted that as tan a = 2 realised that tan b had to be 2 and thus the remainder of the diameter must be 12. A much more efficient and concise method, but not as much fun!


Steve’s method made me realise we are dealing with three similar triangles, and that fact alone is enough to know the ratio of the sides are equal, and hence the rest if the diameter must be 12. This idea was the one used by Rob Rolfe (@robrolfemaths).


Drawing a line from the centre

Jo Morgan (@mathsjem) tweeted this solution:


Which I think is brilliant. Instead of calculating the rest of the diameter, Jo went for the rest of the radius, using a line from the centre to the point the perpendicular touches the circumference. Pythagoras’s Theorem and algebraic manipulation drop the radius out quickly implying the answer in great time.

The I only need Pythagoras’s Theorem method

This was tweeted to me by Mr Draper (@mrdrapermaths):


As you can see, he has used Pythagoras’s Theorem on all three triangles and solved the simultaneous equations to find the necessary information. Possibly not the most concise method, but certainly no less concise than my approach, and one that has a lovely feel to it.

Constructing the geometric mean

Martin Noone (@letsgetmathing) tweeted this fantastic solution:


It is, I believe, derived from an old method for constructing square roots that I think dates back to ancient Greece. This rule is illustrated in hospital top sketch, then he’s enlarged it to ensure the one is a three. He later tweeted the proof, which he uses as an exercise in lessons and one I may start to use too!



Ed Southall then tweeted this pic which shows the general rule and how a semi-circle can be used to construct the geometric mean of two numbers, which itself could be used to find the diameter.


And is a great visual proof that the geometric mean can never exceed arithmetic mean.

Coordinate geometry

Stuart Price (@sxpmaths) informed me that he’d enjoyed the puzzle and left it in his board for his students to complete, and that they used a variety of methods. He mentioned coordinate geometry and equations of circles, so I thought I’d have a crack at that method:


I drew the axes so that the origin was at the point on the diameter where the perpendicular touches it. Ie 3 cm in from the circumference. I considered the general equation of a circle:

(x-a)^2 + (y-b)^2 = r^2

Then subbed b=0 in as the centre lies on the x axis.


I know two points on the circumference, (-3,0) and (0,6) so I subbed these in to get two simultaneous equations which I then solved.


Once I knew a (the x coordinate of the centre) I could work out the radius, and hence the area. A brilliant solution.


Stuart also mentioned calculus, specifically implicit differentiation. I’m not sure how this would help, to be honest, and I’d love to see this solution. I did, however, consider integration. I rearranged the equation of the circle:


I started to integrate but realised I’d need multiple substitutions, so I ran it on Wolfram Alpha instead.








An nice, but incredibly long winded solution!

Nice summation puzzle

December 6, 2014 Leave a comment

Chris Smith (@aap03102) runs a weekly maths newsletter during termtime. If you’re not on the list, I’d advise you to get on it (drop him a line in twitter). It has some great stuff in it, such as this fraction:


How ace is that?

He also has a puzzle of the week feature. This week’s looked cool, so I had a crack at it:


I thought about it a little, and saw that it boiled down to this:


I had the general term: 1/((n^1/2)+(n+1)^1/2) but I wasn’t sure where that would get me. I tried rationalising the denominator of the first two terms:


Now I could see where I was getting to, I thought I’d check the general term, to be sure:


Then it was a case of looking at the sum:


The sum of (n+1)^1/2 -n^1/2 for n=1 to n=91808

Then I cancelled:


Leaving just

91809^1/2 -1^1/2

Which is

303-1 =302

A lovely neat solution to a lovely summation problem which I hope to try in my sixth formers soon, and which I hope will work well with the new GCSE specifications.

Area of a semi-circle puzzle

December 6, 2014 3 comments

If you have read this blog before, you may have noticed I enjoy a good mathematical puzzle ever now and then. This week I had a bit of time to pass and I thought I’d have a crack at this one from Ed Southall (@solvemymaths):


Well almost, I didn’t have my phone and I misremebered it slightly, so actually had a good at this:


As you can see, it’s the same puzzle but scaled by a factor if 1/2.

The first thins I did was label the 4 points. I knew ABD was a right angled triangle, as it is a semi-circle. We are told that ABC and BCD are right angled in the question. The calculate the area we need the radius, which should be easy enough to calculate through Pythagoras’s Theorem and Right Angled Triangle Trigonometry.

I used Pythagoras’s Theorem to calculate AB.


I could then calculate the sine of angle ABC, which I called x. I know that ABC and CBD (which I called y) are complementary, and as such Sin x = Cos y,  so Cos y = 1/rt5


This meant I could easily work out the length BD using the fact that Cos y = adjacent/hypotenuse.


This left me with the two short sides of the right angled triangle ABD, so the hypotenuse (the diameter of the semi-circle) was easy to calculate:


From this the radius, then the area, follow easily.



When I came to type this up I realised I’d solved the wrong problem, so for completeness sake, to solve the original problem I can multiply by 4 (as I need to enlarge so lengths gave increased SF 2, and I’m dealing with area) giving a final answer of: 112.5pi (or 225pi/2).

I enjoyed this puzzle, and did it without a calculator. I think had I used a calculator, it may have lost some of its appeal. I would love to see questions like this appear on the non-calculator GCSE paper. The skills/knowledge needed are circle theorems, Pythagoras’s Theorem, Right Angled Triangle Trigonometry, Surds, Area of a Semi-circle and basic number skills. All of which should be within the grasp of a decent GCSE student. There are, of course, many other solutions, some of which are explored here.

Quadratic Equation Solver – the app

December 3, 2014 Leave a comment

Earlier today Will Davies (@notonlyahatrack) tweeted at me that Phil Boor (@Phil_Boor) had had his first app published, “Quadratic Equation Solver“.

I quickly downloaded it, and I quite like it. It does exactly what it days on the tin, it solves quadratic equations. I’m not sure there’s a huge market for it to be honest, especially with the other maths apps that are available that solve quadratic equations and do lots of other things as well.


One love the look of it, and how quickly it solves quadratics, but I did find it slightly annoying that you have to put 1 in as a coefficient. That you can’t change the + symbols to – symbols so have to add negative coefficients.


I was also a little disappointed that it didn’t give me the formula when I tried to solve ax^2 + bx + c = 0.

It amused me that for this:


Expresses the answer -1 as -3+rt4 and -5 as -1-rt4!

All in all, a lovely little free app, if you want one that only solves quadratic equations. If you’re after one that does this (albeit in a slightly less cool way) and more, try the free “mathematics” or the very reasonably priced “Wolfram Alpha“.

Edublog Awards 2014

December 1, 2014 1 comment

My timehop today informed my that it’s a year since I posted my nominations for the Edublog Awards 2013. I had forgotten they existed and thought I would enter some nominations. Unfortunately, nominations have closed. So this post is who I would have nominated!

Best Individual Blog

This is a tricky category for me, as I love loads and loads of blogs. But after thinking long and hard about it I have decided my nomination for “Best Individual Blog 2015” goes to Colin Beveridge’s (@icecolbeveridge ) “Flying Colours Maths“. The blog is, for me, the perfect mix of maths curiosities and maths teaching ideas, all of which Colin purveys with great humour. He is always interesting and entertaining, although on occasion he is wrong.

Best Group Blog

A blog I’ve discovered recently which brightens my day is “One good thing“. It’s run by a number of people, including Sam Shah (@samjshah) and Tina Cardone (@crstn85). It is full of positive reflections and some excellent maths teaching ideas.

Best New blog

This is another tough choice for me, as there are so many new blogs that are brilliant. 2014 has seen a wealth of fantastic new bloggers reflecting and sharing. The choice was really tough, and the only way I could choose from the final two was that the other would also be nominated in another category. So my nomination for “Best Newcomer” is Amir’s (@workedgechaos) “Teaching at the edge of chaos“. This is a brilliant blog that has helped me frame my thoughts on many subjects and has given me a fascinating insight into Amir and many other maths teachers through the behind the mathematician series.

Best Resource Sharing Blog

Who else could this one go to but the Fabulous Joanne Morgan’s (@mathsjem) “Resourceaholic“. It’s a phenomenal blog, dedicated to sourcing the best resources for maths lessons. Special mention goes to the weekly “Maths Gems” series.

Best Teacher Blog

A strange category, as all the above blogs are written by teachers. In thinking about this category I wanted to choose a reflective blog that puts things into perspective. For this Is have chosen Mel Muldowney’s (@just_maths) superb blog, which has been particularly poignant for me as I too have changed schools this year and met many similar challenges.

Best Headteacher Blog

There are some fantastic headteacher blogs put there, that always make me think. This year I have found quite a few of them to be particularly inspiring, and I’d have liked to nominate Tom Sherrington’s (@headguruteacher) “Head Guru Teacher“, I’ve loved reading about Tom’s journey from KEGS to Highbury Grove.

Most Inspirational Post

This year, I think the most inspiring post I’ve read was this from John Tomsett (@johntomsett). It sets out a vision, a mission statement, of what a great head should be.

Lifetime Achievement Award

This year I’d have to nominate Old Andrew (@oldandrewuk). For his incredible and tireless work on “The Echo Chamber” coupled witg his own blog “Scenes from the Battleground“. I don’t always agree with all he writes, but sometimes he’s just spot on.

Now that nominations have closed, you can vote for the Edublog Awards here.

If you are interested on seeing a more comprehensive list of blogs,and pages I like, click here.

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