Chelekmaths
Saw this great question on the 2016 UKMT Senior Maths Challenge and wondered whether it would be possible to make some of my own. I had just been teaching Surds and wanted some nice enrichment that linked surds to indices.
I realised that you can nicely create your own by working backwards from the answer (See Below)
So I created a few more!
You can download the PPT here:
Can you try making some yourself?
I have said before on this blog that I absolutely adore the UK Maths Challenge as a way of massively improving your problem solving skills and using preexisting tools in totally new ways. It’s also an absolute goldmine for interesting purposeful practice. One of my favourite way of creating resources is by choosing a UKMT (UK Maths Trust) question that I particularly enjoy and seeing in what glorious directions it can lead.
Lets look at the question above – what seems like a relatively simple question has a wonderfully elegant solution. I will include the answer at the bottom of the post – although as usual I would recommend trying the question first!
So to start with students are encouraged to just try the question themselves – they have the whole page to write down their thoughts and try different things and if they want some support they can turn to the next page.
They can work through the simpler example I have given them and then go on to doing very similar questions to the original so they can practice the steps involved. These slight variations mean that students can use what they have used in previous examples to help when the questions get slightly weirded!
The questions progress, using an differing increases before finally looking at a more general example!
Although the question feels pretty niche there is lots of maths to be practised here. The link between a percentage increase and the corresponding numerical multiplier, solving simple quadratic equations and some nice thoughts on fractions lead to an exercise that practises some key skills in a nice creative way! I plan on using it as some problem solving practice after having taught part of those topics.
Please feel free to try it by downloading here – The answers are available through the resources under the subheading “Problem Solving”
How would you attempt the above question? You could just work out the multiplication and then square root that product to find out the missing value. But square rooting in notoriously a tricky function without a calculator and also where is the fun in that! It also will become an increasingly impractical method the larger the numbers get – for example in the question below…
As with lots of the worksheets I make I like to start with simple questions for students to work through and throwing in an example to help but the aim is for it to be a relatively self-explanatory exercise that slowly builds to the intended questions that we want to solve!
These questions also came from a wonderful UKMT question that I saw and needs some knowledge about the structure of square numbers to answer it. In particular that the prime factors of square numbers have to come in even pairs. For example 16 can be represented as 4 x 4 but also as two pairs of 2 x 2. The number 18 on the other hand is built from 3 x 3 x 2 so you have a pair of 3s but then one factor left over so it is not a square number. How could we make 18 a square number just by looking at its factors? We need to find a buddy for that extra factor of 2 so by x 2 we end up with 3 x 3 x 2 x 2 = 36 which is a square number! You can even rearrange it slightly and write it as (3 x 2) x (3 x 2) to make it even clearer that it is a square number as it is something times itself.
So for the question above you can rewrite 64 as 8 x 8 and 4 as 2 x 2 and by rearranging you then have (8 x 2) x (8 x 2) which means that 64 x 4 has to be 16 squared! A wonderfully simple way of solving these questions and also a great excuse to practise factorising (prime and not prime!)
You can try the worksheet yourself here and the answers can be found on our resources page. Its such a delightful little puzzle with a lovely solution – Even coming up with your own questions is fun so I implore you to try that too!
The UKMT is an absolute goldmine of maths and puzzles. Even more excitingly the UKMT has extended solutions to their maths challenges that include additional questions and investigations for each question so I would definitely suggest checking that out here
Below is the solution to the original problem:
As always thank you so much for reading and if you are interested in getting involved with us let us know here and you can also subscribe so you don’t miss out on any content by entering your email below!
NJK
Firstly I would urge you to try the question in the photo yourself. It requires some knowledge of powers and solving quadratics and is wonderfully satisfying to find all SIX solutions.
Questions like this are a great example of finding creative ways to practice fundamental skills. In this case one of the key skills being practised here is the ability to solve quadratic equations and rearranging them into a form that you can solve easily. So how to introduce this to a class?
Lets look at the example in the picture – How can we can get to a solution of 1 on the LHS of our equation? The two cases that make this happen: Case 1 by setting (x-1) = 1 and case 2 by setting (x-1) = -1
Both these methods when squared will give us our answer of 1 and so that is where we will get our final values of x to be 2 and 0.
So taking it a step further we now an expression as our power which gives us a third case: Setting x+2 to zero. When the power is zero whatever is inside the bracket is irrelevant as any real numbers to the power of zero equal to 1.
Okay 1 to the power of anything is one so x-1 = 1 gives us our first solution: x=2
In case two we need -1 to an even power to get to one so we have x-1 = -1 which gives us a solution of x=0 which when put into the power gives us an even power so it must be a solution
Finally in case 3 we set x+2 = 0 which gives us our third solution and final solution which is x=-2 and it is solved! Now can we use this simpler example to work our way up to the six solutions of the first problem?
I used the sheet above as a starter in my first lesson with my new year 11s – as a little bit of a fun way to gauge both their ability and their desire to apply maths they definitely know in ways they might not have seen before. The questions don’t lead onto each other as nicely as I might like but once we had discussed the first few the whole class were able to get their teeth into the rest of them.
By working through this small exercise they were able to practise factorising and solving quadratics in a way that felt fresh and interesting whilst also giving me an opportunity to circulate and learn more about my class.
I will not explain the full explanation for the question as it seems only fair to direct you to the original video here for a lovely detailed work through of the solution. Once you have watched the video try the rest of the worksheet!
I encourage you to try each of the questions looking at each of the three cases. Then try and come up with some of your own that fit this format – I always think one of the best ways to improve is to play around with writing your own questions and these are a fun way to start!
If you want to download the worksheet as a word document click here
Answers can be found on our Resources page
Thanks for reading!
NJK
I recently came across this fencing problem posed by two mathematically minded farmers, the first part of which I feel is rather well known. Questions such as these are easy to visualise, meaning you can often give an instinctive answer and then test if your intuition was correct after solving the problem.
Consider an established farmer who has a very long (infinitely, if need be) straight fence forming part of the boundary of a field. Generously they have allowed you to use this fence, while constructing your neighbouring plot of land. Therefore we plan to enclose a rectangular field, only needing to build 3 sides of the quadrilateral. We will stipulate that the maximum length of these 3 sides is a length of L. The neighbours fence is shown below, on the left vertical.
Question: Using a fence of total length L, what ratio of sides will give the maximum area?
The idea here is to use the constraint on the fence length L to create an equation using our unknown side lengths, x and y. We can then use this to create a formula for the Area, using the constant L and one unknown.
With this formula in place, the problem then becomes a question of maximising the equation, which we can achieve by finding the stationary point of the derivative.
I’ve solved this below, so don’t scroll down if you would like to try this yourself!
I’ve also created a Desmos Graph to see how the area changes with different values of L.
So the answer is a 2:1 ratio in the direction of the neighbours fence. While this may not have been your initial intuition, it follows that adding to the fence in the vertical direction only costs half as much as adding in the horizontal direction, due to the free fence offered by our neighbour. Therefore, the maximal area will be achieved by having a vertical length double that of the horizontal length.
We will leave how our generous neighbour tends to their field of infinite size to another day.
Considering the success of the first project, we have decided to expand our farm by building a new fence for a new field. This time the plot of land is somewhat different.
In this case we have a square farm building in the middle of another very large field. We’ve realised that if we include the edge of the building as part of the fence, we can increase the area we enclose.
One option is to align the field with the edge of the farm building (shown on the left below). However, our established neighbour has suggested we might want to consider using the long diagonal of the building as part of the fence. These two scenarios are shown below for the vertical and diagonal configurations building.
Question: Again considering a total fence length L and a square building of length l in both scenarios, which configuration gives the maximum area?
Again the solution to these two configurations follows a very similar pattern to the warm up question above. However, instead of working out the ratio of the side lengths, we need to find the maximal area of each scenario.
In the interest of time, we will solve these in parallel as the solutions are very similar. Additionally, there is a neat trick that we can use to minimise the effort required. In opening two statements, we can observe that both equations have equal coefficients of x and y. Therefore, we can interchange x and y freely in the formulas (since they are variables we have made up to understand the problem). This means any solutions we find for x, is automatically valid for y and is called a symmetric argument.
Now all that remains is to determine which of the two areas in the final row are larger. If we assume that the vertical configuration yields a larger area than the diagonal configuration, we can create an inequality to simplify in terms of L and l.
So we have found that the relative lengths of the fence and the side of the building impact what choice we should make. If the fence is much shorter than the side of the building, then we should choose the vertical configuration. Of course, we can reverse the inequality to find the alternate scenario.
Again, I’ve created an interactive Desmos Graph that gives a visual interpretation of the problem.
With the two solutions in place, we can now compare against our original instincts to check if we were correct. If you were, well done! If, like me, you were not, I find it useful to examine to think about why my intuition failed in order to improve for next time!
And finally, as the old saying goes, the enemy of my enemy is my fence.
SED
Magic squares always feel like a go-to end of term fun activity and one I have never given a particularly large amount of thought to – when I saw this question come up on a facebook maths page I am a member of I wasn’t sure how easy it would be but I talked it through in my head and came to a nice answer (I will go through the answer at the end of this post – Try it out yourself!) and I knew the reason I was able to solve this puzzle so easily was thanks to my new favourite YouTube channel: Cracking the Cryptic
It’s hard to describe the pure joy that the YouTube channel crackingthecryptic has given me over the lockdown period. Glorious and regular puzzles that have honestly captivated me for hours on end. For a few weeks solving their latest puzzle over breakfast (and inevitably lunch because they were so tricky) became a minor addiction for me and lead me to many great Aha moments that my housemates soon tired of hearing about!
The above puzzle makes use of a simple 3 by 3 magic square in the middle of the grid that must contain all the digits 1-9 with all rows/columns/diagonals adding to the same amount. But how to even get started? As you know that all together the sum of the numbers 1-9 is 45 that means that each row/column/diagonal has to contain numbers that sum to 15 so that the three rows/columns add to the same amount and together add to 45. Then you can work out what number has to go in the middle an therefore is part of every single row/column/diagonal – the number ends up always being the median of the numbers you had to start with – so in this case it is 5 and that’s your starting point for the rest of the puzzle!
It was this previous knowledge that helped me to answer the question at the top of this post so quickly – because I had seen similar problems in different contexts I was able to use those problem solving skill to quickly find the number that should go in the middle square and then the rest of the problem solves itself from there. The more maths/puzzles you play around with the better you will be in the future!
Below are screenshots and links to some of the bests puzzles I have attempted and sometimes even completed. I urge you to try them yourselves and let us know what you thought.
The more puzzles and problems you grapple with the better you will be at grappling. Also it’s shocking how rarely I actually sit down and concentrate on something for an hour at a time without distractions and these Cracking the Cryptic puzzles gave me many of those hours.
How to solve: You know that as the whole grid must contain the numbers 7-15 the total of everything within the grid must be 99. As there are three rows/columns we divide by 3 to get 33 and therefore each r/c/d must equal 33 so the top left box must be 33-14-7=12
We also know that the number in the middle is the median number from 7-15 so must be 11 and then we fill in the rest of the grid from there! “n” must be 33-11-14 so n = 8
Let us know if you try and of the puzzles by clicking here – Thanks for reading!
NJK
These three seemingly very different questions all showed up in the same SMC paper in 2017. Try each one yourself and see if you can work out what one trick links them all together!
I’ve always been thrilled by the ability of using the “Difference of two squares” to simplify problems. Whether its answering numeracy problems quickly and easily like the one above or using it to crack open harder problems like the three SMC problems above.
At the bottom of this post I will include some worksheets (not made by me) that practice using this skill in simple and fun ways as well a lovely geometric proof of the above formula. There will also be some extension questions as well as seeing how these techniques show up in the A level curriculum.
This question is a great example of just starting off trying to simplify your answers and seeing what pops out. Looking at the multiple choice options available here its not immediately clear how any of those could possibly be the correct answers.
The most interesting part of working through this is when you get to the following step 
Recognising that by definition 1 is a perfect square means that whenever you have a square number mius one you can use this formula. This give you the final piece of the puzzle which simplifies to answer E as shown below. Such a elegant question!
This one has a longer but no lesson elegant solution. There are plenty of ways to solve this one but the words “The difference between the squares of their ages” should immediately get the DOTS alarm bells ringing. Using DOTS simplifies this problem down in a lovely way and makes the algebra much more manageable as the answer below shows.
There is also some extension to this question which are a great way to practise the skills yourself – try them yourself and contact us if you have questions/answers!
It is not immediately clear how DOTS could help you with this question. The first thing to say is that the main reason I thought to use this trick is simply that I have answered lots of problems before. There is no substitute for practise and increasing your mathematical toolkit but playing around with lots of different maths. Needing almost any excuse to use DOTS and additionally y squared in the question meant that I went down that route first of all.
I’m not going to go over every step of the question but the step where you set each bracket to different factor pairs of 15 is not an easy step to see – once again there is no replacement for practise and seeing this question will help with similar questions in the future – I will also include an A Level question that uses this exact technique later on.
Honestly an absolutely glorious question. It is no surprise that it is one of the last questions in the Senior Maths Challenge (For school years 10+) when it requires lots of quite difficult techniques (also under a timed exam pressure! Very tricky!)
This A level question is a lovely way of seeing how theses skill show up in the school curriculum and are a great way to prove some seemingly not straightforward mathematical statements. Try it yourself! I have attached a link below to a document with a group of these questions and importantly the answers to them.
Difference of two squares is a wonderful trick that has a multitude of problem solving uses. The more problems you do and the more you can recognise this the easier (and more fun!) these questions will become.
I have included below a link to worksheets created by the legend Don Steward that can help you practise DOTS in numerical settings.
If you have any extra thoughts/questions please do send us a message and we will get back to you as soon as we can!
NJK
I was recently given this problem by a colleague and it immediately took my interest. As soon as I saw it my first thoughts were not “what is the answer?” but “is there an answer? Is there more than one answer? Can I prove this?”.
I spent that lunchtime in my classroom looking at the problem. I started by looking at the digits in the units column and noticed that since the value of the units in the answer must be z then x + y = multiple of 10. This gave me different possibilities to consider:
This was the first breakthrough in the problem as I now had an equation that places a constraint on the possible values of x and y.
I then adopted a similar approach with the tens column (x + y = 10 so don’t forget to “carry” the 10 into the tens column which I initially did!). Looking at the digits we can see that 10x + 10z + 10 = 100x which we can simplify to get z + 1 = 9x. Now since x, y and z are all single digit numbers I could then use the two constraints to deduce their values:
x + y = 10 (A)
z + 1 = 9x (B)
So I had obtained “the answer” and could see that it works. It’s always satisfying to find the answer to a maths problem but it’s even more satisfying to say with confidence that this is the only answer and prove it with reason. I presented this problem to a high ability year 8 class with no instructions and was intrigued to see how their thinking was to trial numbers and simply search for “the answer”. As expected, many of them were able to find the correct values for x, y and z but weren’t able to tell me with any conviction if these are the only possible answers.
I am sure there are other proofs to this relatively straightforward problem and I would be interested to see people’s suggestions!
DTH
Chelekmaths 
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