# Multiple Surprises

*Multiple Surprises printable worksheet*

Here are some challenges involving consecutive numbers and multiples.

**three consecutive numbers**where the first is a multiple of 2, the second is a multiple of 3 and the third is a multiple of 4?

Can you find several examples?

What do you notice?

Can you explain your findings?

What if the first is a multiple of 3, the second is a multiple of 4 and the third is a multiple of 5?

What if the first is a multiple of 4, the second is a multiple of 5, and the third is a multiple of 6?

Is there a way to find sets of

**four consecutive numbers**which are multiples of 2, 3, 4 and 5 (in this order)?

Or

**five consecutive numbers**which are multiples of 2, 3, 4, 5 and 6 (in this order)?

Can you use what you have discovered to help you find a few sets of ten consecutive numbers in which:

- the first is a multiple of 1
- the second is a multiple of 2
- the third is a multiple of 3
- the fourth is a multiple of 4
- the fifth is a multiple of 5
- the sixth is a multiple of 6
- the seventh is a multiple of 7
- the eighth is a multiple of 8
- the ninth is a multiple of 9
- the tenth is a multiple of 10?

*With thanks to Don Steward, whose ideas formed the basis of this problem.*

Thank you to everyone who submitted a solution for this problem. There were just too many to mention everyone, but a lot of good work was submitted, so well done all!

Many of you started off with a 'trial and error' approach and then looked for patterns in what you had found:

If I write the first few sequences I found in columns like this:

$2 3 4\\

14 15 16\\

26 27 28\\

38 39 40$

I can see that each number is $12$ more than the number above, and so adding $12$ to each sequence will work.

Some people noticed that adding $10$ to the last number in one sequence and starting the next sequence from there would work, which is also true since this is effectively the same as adding $12$ to each number in the sequence.

Some people then looked into *why* adding $12$ seems to work:

I noticed that $12$ is a multiple of $2$, $3$, and $4$ and, more than that, $12$ is in fact the __lowest common multiple__ of $2$, $3$, and $4$.

That is why adding $12$ works, because adding a multiple of $2$ to $2$ gives a multiple of $2$ and adding a multiple of $3$ to $3$ gives a multiple of $3$ and the same for $4$. So, if we start from the obvious consecutive sequence, $2, 3, 4$ and keep adding $12$ we will get more and more consecutive sequences which fit the criteria.

Hannah also looked for lots of sequences but then tried a slightly different approach:

I found the sequences:

$2, 3, 4\\

14, 15, 16\\

26, 27, 28\\

38, 39, 40\\

50, 51, 52$

In order to generate one of these sequences, I had to find the nth term rule. Because there are three terms per sequence, I decided to add the terms together to give me one number that would represent that sequence.

$2+3+4=9,\\

14+15+16=45,\\

26+27+28=81,\\

38+39+40=117,\\$

and $50+51+52=153.$

So the new terms of the sequence are: $9, 45, 81, 117, 153$.

The difference between each of these numbers is $36$ so my expression for the new sequences is:

$U = 36d + a$

And since the first term in the new sequence is $9$, that means $a=9$ and so:

$U = 9 + 36d$

I checked that my expression for the sequence worked and it did!

Then I tried to work out how to get from my new sequence back to the original sequences of three consecutive numbers.

I noticed that if I did $\frac{U}{3}$ I would get the middle term from my original sequence of three consecutive numbers. I think this is because the middle term in three consecutive numbers is also the average of the three terms.

Hannah is correct! We can show this algebraically:

Let's call the middle term, $p$.

Now, since the three terms in our sequence are consecutive we can write them as $p-1, p, p+1$.

So when we add the three terms together to get the number Hannah has called $U$, we are adding $ (p-1) + p + (p+1) = 3p$.

This is why Hannah can use the $U$ sequence to find new sequences of three consecutive numbers which satisfy the criteria in the question.

For the second part of the question, looking for three consecutive numbers which are multiples of $3$, $4$ and $5$ respectively, most people continued to use the idea of lowest common multiples.

For the numbers $3$, $4$ and $5$, I did the column strategy again, and figured out how the same thing happens, but with the number 60, as that is the Lowest Common Multiple of $3$, $4$ and $5$.

$3, 4, 5,\\

63, 64, 65,\\

123, 124, 125,\\

183, 184, 185,\\$

The lowest common multiple of $4$, $5$ and $6$ is also $60$ and so the columns for those numbers look like this:

$4, 5, 6,\\

64, 65, 66,\\

124, 125, 126,\\$

and so on.

We can prove algebraically why the LCM method works:

Starting from our three consecutive numbers, let's say $a, b, c$ we want to find more sequences of three consecutive numbers which are multiples of $a$, $b$, and $c$ respectively. This means we want

$a + m = $ a multiple of $a$

and $b + m = $ a multiple of $b$

and $c + m = $ a multiple of $c$

Therefore $m$ must be a multiple of $a$, $b$ and $c$ and to make sure we don't miss any possible sequences $m$ must be as small as possible. This proves that $m$ must be the LCM of $a$, $b$ and $c$.

As a lot of you noticed, regarding the final part of this question:

The LCM method will work for $4$, or $5$ or any number of consecutive numbers.

For example Rianna from Dr Challoner's High School for Girls looked at sequences of seven consecutive numbers:

A slightly more challenging one is using multiples of seven consecutive numbers such as $2,3,4,5,6,7,8$. The lowest common multiple of these is $840$.

$2,3,4,5,6,7,8$

$842,843,844,845,846,847,848$

$1682,1683,1684,1685,1686,1687,1688$

And Malory from DCHS looked at sequences of nine consecutive numbers

I then wanted to challenge myself further! I used $2, 3, 4, 5, 6, 7, 8, 9$ and $10$! I tried to find nine consecutive numbers and I found out that the lowest common multiple is $2520$. I added $2520$ to the nine consecutive numbers and it made a set of consecutive multiples.

Well done again to everyone who had a go and submitted solutions!

### Why do this problem?

This problem offers a great opportunity for students to take the important mathematical step of starting from particular cases and moving towards generalisations. In exploring each of the challenges, students will get plenty of practice at working with multiples and lowest common multiples, which will deepen their understanding of factors and multiples and help them appreciate the significance of prime factorisation.

### Possible approach

*This printable worksheet may be useful: Multiple Surprises. This problem featured in an NRICH Secondary webinar in October 2020.*

Ask students the first question:

"Can you find **three consecutive numbers** where the first is a multiple of 2, the second is a multiple of 3 and the third is a multiple of 4?"

Give them some time to think and then ask "Can you find another set? Can you find a set that you don't think anyone else in the class will have thought of?" If students have whiteboards they could write a set on their boards and hold them up. You could collect sets of three numbers that satisfy the constraints on the board for everyone to see.

"What do you notice about all the sets we've found?" Give students time to think on their own and to discuss with a partner, before discussing it with the whole class. Students might notice that the middle number is always an odd multiple of 3, and that there is a gap of 12 between one set and the next.

You could also invite students to explain how they found examples - this is a good opportunity to emphasise the importance of working systematically when tackling a maths problem.

Then set the next two challenges and invite students to work on them in a similar way:

"What if the first is a multiple of 3, the second is a multiple of 4 and the third is a multiple of 5?"

"What if the first is a multiple of 4, the second is a multiple of 5, and the third is a multiple of 6?"

Once they have had a chance to find some examples and make some conjectures, bring the class together to share findings, and discuss any conjectures: "When we were looking for multiples of 2, 3 and 4, we added 12 to find the next set of solutions. When we looked at 3, 4 and 5, there was a gap of 60 between solutions, and there was also a gap of 60 when we looked at 4, 5 and 6. Can anyone explain why?"

Students may explain that they need to add a multiple of 3 to get another multiple of 3, a multiple of 4 to get another multiple of 4, and so on, so for every number in the set to work, they need to add a **common multiple**, and to find the next possible set they need the **lowest common multiple**. Once this insight has been understood, students can try the other
challenges:

"Is there a quick way to find sets of **four consecutive numbers** which are multiples of 2, 3, 4 and 5 (in this order)? Or **five consecutive numbers** which are multiples of 2, 3, 4, 5 and 6 (in this order)?"

Finally, the last challenge offers students the opportunity to show that they have really understood the structure of the problem:

"Can you use what you have discovered to help you find a few sets of ten consecutive numbers in which:

- the first is a multiple of 1
- the second is a multiple of 2
- the third is a multiple of 3
- the fourth is a multiple of 4
- the fifth is a multiple of 5
- the sixth is a multiple of 6
- the seventh is a multiple of 7
- the eighth is a multiple of 8
- the ninth is a multiple of 9
- the tenth is a multiple of 10?"

### Key questions

If I add the same number to a set of three consecutive numbers, will the new set of numbers be consecutive?

If I know that a number is a multiple of 3, what do I need to add to it to get another multiple of 3?

Which numbers are multiples of 2, 3 and 4?

### Possible support

Encourage students to start with just a pair of constraints. For example, "Can you find two consecutive numbers where the first is a multiple of 3 and the second is a multiple of 4?" Then once they have found a family of examples, add the third constraint.

### Possible extension

Invite students to consider how they can extend the last part of the question to generate a set of $n$ consecutive numbers so that the first is a multiple of 1, the second is a multiple of 2, the third is a multiple of 3 and so on. They might also like to consider how this relates to the idea that it's possible to find long sequences of consecutive numbers that do not contain any primes.