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Worked GCSE Number Questions

A problem with factors

A problem question I tackled with a student yesterday asked..

What is the smallest number that has 1, 2, 3, 4, 5, 6, 7, 8 and 9 as a factor

I had just taken along a pack of ‘challenging questions’  because he is an able student but needs a challenge. As is happens, this is a question about Lowest Common Multiple, and I had been looking at this a day earlier with another student.

Anyway, yesterday’s student was quick to point out that we can ignore 1 because ‘1 goes into everything’. That was a good start!

He then started listing all the numbers 9 goes into – the 9 times table – and checking off each number against 8, 7, 6 and so on…  When I showed him the quicker way I’m sharing below here, he admitted ‘That would have taken a long time”

The quicker method is to split all the numbers into their prime factors

2 = 2;  3 = 3; 5 = 5; 7 = 7 – thats the prime numbers in the list

4 = 2 x 2;   6 = 2 x 3;   8 = 2 x 2 x 2;  9 = 3 x 3

Now,  consider the answer we are going to get.  As I usually do I’ll give this a letter, but for a change I’ll call it Z!

Z is just a number, even if we don’t know what it is yet. so we will be able to write Z as prime factors, and, those Prime factors are going to be a bit like the prime factors we’ve already found for all the numbers.

We could find Z by combining all the prime factor lists we have, but we will find we don’t need ALL of them.

For example 8 = 2 x 2 x 2  – and we find we don’t also need to add in the 2s from the 4 and the 6! With the 2s in the 8, we have them covered.

Taking that for all the numbers, for each prime number in our list, we only need as many as covers the most in any number.

2 x 2 x 2 – because 8 is the number with the most 2s

3 x 3 – because 2 is the most number of 3s in the prime factors of our numbers

then 5 and 7, because they only appear once in any list

The answer is – 2 x 2 x 2 x 3 x 3 x 5 x 7. You can do this on your calculator and get 2520 – The answer to this question)

[Actually, I try not to use a calculator unless I have to; keeps my mind sharp! Instead I pick numbers from the list that make multiplying in stages easier.
Pick the 2 x 5 = 10.  The 7 x 3 x 3 = 63 – Gives 630. Then double this twice for the other 2 2s  – 1260  then 2520!]



Drawing our own map

In my last post I describe how we could get information about distances from a map using a scale. In this post we will look at how we can use the idea of a scale to draw a map of our own

Think of the following question : Source, my own imagination but I have seen similar questions in Exam papers

My local park is 150m long and 120m wide. We need to plan the park to have one football pitch (100m by 60m), a play area for younger children (20m by 20m) and places to plant 5 trees. Each tree should be at least 10m from the edge of the park and 20m from the football pitch. There should be 10m at least between each item.

We are given a sheet of paper measuring 25cm by 20cm.

See if you can have a go. The answer will be posted tomorrow.

So how do Maps work?

Since I’ve spent a weekend studying maps, I thought a quick post on the maths involved would be in order. Although it might not be immediately obvious, what we are looking at are Ratios – and this can be a pleasant change from mixing paint which is what many questions in Exams about ratio seem to be about!

Now here is a map of one of my favourite places in the world – and comments from anyone who agrees are welcome!  The point is, a map represents the place you want to visit so obviously has to be a lot smaller than the place itself!  (Ok so this reminds me of a favourite Blackadder joke but lets not get off the point)

The map shown here is to the scale 1:25000 if you have it before you. (I can’t make a similar claim from the picture you can see, as that will depend on the size of your browser!)

If you measure on the map that you are 2cm from the car park and pub*, how far do have left to walk?

2 * 25000 = 50000cm.  Which is great but we don’t usually measure walking distances in centimetres. 50000cm = 500m or 0.5km

Activity  : Find a map of your town and measure the distance to a place your often visit.  Don’t assume your map has the scale 1:25000.  The scale for your map will be written somewhere, perhaps even the front of the map.

*Actually the first pub I ever had a pint of beer!

Answer to the Number machine Question

Here is a confession – If anyone read my last blog post they will have seen a mistake with the question I posed at the end – If you are looking now, this has now been corrected!

The question asked to fill the second box in.  The first box says  x 2 , so if we feed in 6, the number in the middle is 12.   So the second box needs to be an operation that gets from 12 to 6.


Without the clue now added, you could have at least two different answers.   “Take 6” is what I expected, and with the clue that is now the RIGHT answer.  Without the clue, ‘Divide by 2’ would have worked too.

(So would ‘add -6’ if we are getting pedantic, though really thats the same as subtract 6. Another possible answer would be ‘Raise to the power of 0.721, but I wouldn’t expect people studying Number machines to spot that.  There are probably a lot of other operations that get from 16 to 6 in we go that deep!)

All is revealed with regards to dogs – and sheep!

I gave the dog problem from 2 posts again to my Wednesday student yesterday and he solved it in the way it was intended to be solved.

We start by choosing a letter to stand in for the answer we want. Let S be the number of small dogs. Also let L be the number of Large dogs. We have not be asked to find the number of large dogs but this is part of the situation.

So we can say S  = L + 36 – because there are 36 more small dogs than large dogs. Actually L = S – 36 is the same and will lead us to the answer required more quickly.

Also L + S = 49 – the number of dogs. This ‘equation’ I think is intended – but I will return to this point.

We can solve the equations by substituting the first into the second to give  S – 36 + S = 49.

Simplify by adding the S and the S and adding 36 to both sides, we get 2S = 85   and so S = 42.5

“But how can we have half a dog”, asked my Wednesday student, and a very fair question two. This is the more obvious reason why this is a bad question – interestingly bad but bad none the less. If we are going to encourage students to take ‘real life’ problem solving seriously, then the questions we ask should make sense.

But my other reason why I’d want to change this question comes back to the story about the black sheep.  The point of the  story is that as mathematicians we shouldn’t assume anything – or at least we should qualify any answer by stating clearly which further assumptions we have made – I claim the answer to this question is incomplete unless we also say

‘Assuming all dogs are large or small’  – i.e  there are no medium sized dogs!  Without that, we can’t use safely the equation S + L = 49!



A short trip to Scotland

There is a story about mathematicians that I think I first read when I studying for A-level.


An astronomer, a physicist and a mathematician are on a train in Scotland.  The astronomer looks out of the window, sees a black sheep standing in a field. “All the sheep in Scotland are black!”, comments

“Oh no” says the physicist. “Only some Scottish sheep are black.”

This is the cue for the mathematician to get involved. “No,” he tells his friends. “In Scotland, there is at least one sheep, at least one side of which appears to be black, some of the time.”

I remembered this story yesterday when posting the problem about the dogs, which I reckon wasn’t written by the same mathematician. I’ll explain why in my next post, where I will also give the answer!

How many small dogs? A maths problem from the Independent

Sometimes the time I spend scouring Facebook pays off, and I find a maths problem posted with the a discussion that follows.

This one comes from the Independent’s Website and the story that goes with it says its for a 7 year old,  I’d be interested to see a seven year old tackle it – I have no doubt that a bright one could!  I’m going to share it will my year 10 (15 year old) student this afternoon, and I’ll report back tomorrow, with an answer!

Rules for spotting factors

We often need to spot the factors for a higher number. How can we do this without doing the division sums?

For some numbers, its easy to spot. How can you tell if 5 is a factor of your number?

That’s easy. If the last digit of the number is a 0 or 5, then 5 is a factor. If it doesn’t, then 5 is not a factor

5 is a factor of 670 and 1225. It is not a factor of 234 or 1352.

Is 2 a factor? Thats easy in a similar way. The last digit needs to be a 0, 2, 4, 6 or 8.  The even numbers, of course. 2 is a factor of even numbers. Its not a factor of odd numbers.

Is 3 a factor?  Or 9?  For these possible factors we don’t just look at the last digit, but we add up all the digits in the number. If they come to a known multiple of 3, then 3 is a factor. If they come to a known factor of 9 then 9 is a factor.

So 4524 – add up the digits  4 + 5 + 2 + 4 = 15 – 15 is 3 x 5 so 3 is a factor of 4524, but 9 is not.

But 4527  4 + 5 + 2 + 7 = 18  and 9 is a factor of 18, so it is also a factor of 4527.

Note the pattern also holds for 15 and 18 – I used our knowledge of 3 x 5 = 15 and 2 x 9 = 18 above,  but also 1 + 5 = 6 and 1 + 8 = 9.


How do I know if 6 is a factor?  use both riles above.  If 2 is a facor and 3 is a factor, then so is 6.

What can we really tell from a graph?

A break from the world cup posts today, because I’ve just come across an amusing website.

Correlation – or the lack of it – it one of the most interesting things we can find from statistics.  For years, whether smoking causes cancer was ‘controversial’ but when the results of cancer cases versus smoking were plotted a direct correlation was found.

OK, so there is a small lag in time, but that is explainable..


And that is the point about correlation on a graph. A pattern we can see only means two things are ‘correlated’ if we can explain why that connection might exist.

Its an important lesson in how Statistics must be handled!

And that brings me back to the amusing website, which gives a few comical examples where it LOOKS like there is a connection….   but can there be, really?

Here is an example!









And here is a link to some more!



Coming on in leaps and BOUNDS

For this post, you need to know your basics in rounding numbers. Here is a recap

  1. Pick out the digit in the last place that will survive the rounding process


  2. Look at the next digit… Is it a 0 1 2 3 or 4? If so take no action. Otherwise raise the underlined digit by 1. In this case a 9, so raise to 4
  3. In both cases, remove all following digits after the underlined one


Now, for this subject of bounds we are going to look at this kinda back to front. We are told a number has been rounded to 2 decimal places. So what could that number be?

2.335 is the smallest number it could be. Any number smaller would be rounded to 2.33
2.34499999 is the largest number it could be. Any number larger would be rounded to 2.35.

2.335 and 2.3449999  are the upper and lower bounds to the number given the information we have.

Let’s look at something a bit more complicated

Now, if we know that I have ran 10.6m to 1 d.p  in  19.5 seconds to 1 d.p, then what is my highest possible speed and lowest possible speed, to 3 d.p ?

We are being asked for the ‘upper bound’ and ‘lower bound’ for the speed.

Remember  Speed = distance / time


Start by writing down the upper bound’ and ‘lower bound’ of my distance and time.

Upper Bound of distance is 10.6499m
Lower Bound of distance is 10.5500m
Upper Bound on Time is 19.5499 seconds
Lower Bound on Time is 19.4500 seconds
Now, this is where we have to think because to find the upper bound on the speed, we need the highest possible time BUT the lowest possible time. This is because as time goes up, speed comes down, for any given distance. I use the highest possible value for time in the lower bound calculation, for the same reason

So upper bound on Time = 10.6499/19.4500 = 0.547553

The lower bound on time = 10.5500/19.5499 = 0.539645

The bounds for the speed are [0.540, 0.548]  to 3 d p

Note that I used 4 d.p. in my calculations. I will always use one or more decimal places than required then round at the end. We risk losing accuracy otherwise.