# Odds, Evens and More Evens

Alison, Bernard and Charlie have been exploring sequences of odd and even numbers, which raise some intriguing questions...

*Odds, Evens and More Evens printable worksheet*

Here are the first few sequences from a family of related sequences:

$A_0 = 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29...$

$A_1 = 2, 6, 10, 14, 18, 22, 26, 30, 34, 38, 42...$

$A_2 = 4, 12, 20, 28, 36, 44, 52, 60...$

$A_3 = 8, 24, 40, 56, 72, 88, 104...$

$A_4 = 16, 48, 80, 112, 144...$

$A_5 = 32, 96, 160...$

$A_6 = 64...$

$A_7 = ...$

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.**Which sequences will contain the number 1000?**

Once you've had a chance to think about it, click below to see how three different students began working on the task.

Alison started by thinking:

Bernard started by thinking:

Charlie started by thinking:

Can you take each of their starting ideas and develop them into a solution?**Here are some further questions you might like to consider:**

How many of the numbers from 1 to 63 appear in the first sequence? The second sequence? ...

Do all positive whole numbers appear in a sequence?

Do any numbers appear more than once?

Which sequence will be the longest?

Given any number, how can you work out in which sequence it belongs?

How can you describe the $n^{th}$ term in the sequence $A_0$? $A_1$? $A_2$? ... $A_m$?

For Alison's approach:

What happens to the numbers as you go down the rows?

What happens as you go up the rows?

For Bernard's approach:

Which numbers end in a 0 in row $A_2$?

Which numbers end in a 0 in row $A_3$?

Which of these sequences will hit 1000?

For Charlie's approach:

Can you find a similar method to Charlie's to describe the other rows?

Which descriptions include 1000?

Lots of students gave the correct answer to this: 1000 appears in the sequence $A_3$, and nowhere else. Let's have a look at why!

Andrea from Highgate School noticed that the numbers got very big very quickly, and so it wasn't too hard to write the sequences out:

$A_7 = 128, 384, 640, 896, 1152, \dots$

$A_8 = 256, 786, \dots$

$A_9 = 512, 1536, \dots$

$A_{10} = 1024, \dots$

and after this every sequence starts with a number bigger than 1000. So 1000 doesn't appear in $A_7$ or above. Then we can check all of the sequences $A_0, \dots, A_6$ individually, noticing that the common difference between successive terms in each sequence is constant: in $A_0$, you add 2 to get from one term to the next; in $A_1$, you add 4, etc.

Looking at common differences is a good idea. Sam and Robbie, again from Highgate School, also noted that these sequences had common differences. Robbie wrote:

Sequence $A_1$ goes up in 4s, but starts at 2. The closest it gets to 1000 is 998 or 1002, because 4 is a factor of 1000, so 1000 isn't in this sequence.

Sequence $A_2$ goes up in 8s, but starts at 4, and 8 is a factor of 1000, so 1000 isn't in this sequence either.

Sequence $A_3$ goes up in 16s, and starts at 8; but 16 goes into 1000 exactly 62.5 times, and 8 is half of 16, so 8 + 16$\times$62 = 1000. So 1000 is in this sequence.

Jacob from Highgate School did something slightly different:

In $A_2$, each number is 4 less than a multiple of 8, so if 1000 was in $A_2$, then 1004 would have to be divisible by 8, which it isn't.

In $A_3$, each number is 8 less than a multiple of 16; and 1008 is a multiple of 16, so 1000 is in $A_3$.

Was there another way to do it? Lucas from Elm Grove Primary School did it this way:

The $A_0$ sequence is every odd number times 1.

The $A_1$ sequence is every odd number times 2: 1$\times$2, 3$\times$2, etc.

$A_2$ is every odd number times 4: 1$\times$4, 3$\times$4, etc.

$A_3$ is every odd number times 8.

It carries on this way: every sequence is all the odd numbers times a certain 'multiplier'.

So if 1000 is in one of these sequences, it must be an odd number times one of these multipliers.

The multipliers are 1, 2, 4, 8, 16, and so on. So, for example, if 1000 was in $A_2$, we would be able to write it as 4 times an odd number.

We can write 1000 as 1$\times$1000, 2$\times$500, 4$\times$250, 8$\times$125, 16$\times$62.5, and so on.

1000, 500 and 250 aren't odd, and 62.5 is a decimal (and if we keep going we'll keep getting decimals).

But 8$\times$125 is of the right form: it's the multiplier for sequence $A_3$, times an odd number. So 1000 only appears in sequence $A_3$.

A good explanation! We also got lots of similar solutions from Connor from Forest Lake State School, Ben from High Storrs, Zoreb and Noor-ul-Ain from Westfield Middle School, Jacob from Highgate School, Holly, Harry and Simon from Beer C of E Primary, Joe from Colchester Royal Grammar School and an anonymous student from Cage Green Primary School. Well done!

Some students started by calculating the n-th terms of the sequences. Faisal, Florence, Kimen and Siobhan from Strand on the Green Junior School gave this fantastic explanation:

The n-th term in each sequence is:

$A_0$: 2n - 1

$A_1$: 4n - 2

$A_2$: 8n - 4

$A_3$: 16n - 8

$A_4$: 32n - 16 etc.

That is, the n-th term in each sequence is double the n-th term in the previous. So:

$A_0$: 1$\times$(2n - 1)

$A_1$: 2$\times$(2n - 1)

$A_2$: 4$\times$(2n - 1)

$A_3$: 8$\times$(2n - 1)

$A_4$: 16$\times$(2n - 1) etc.

This is like a sequence of sequences! We have the sequence 1, 2, 4, 8, 16, ... appearing. So we thought about how we could calculate its m-th term. We soon noticed that this sequence was exactly $2^0, 2^1, 2^2, 2^3, 2^4, \dots$. So:

$A_0: 2^0\times (2n - 1)$

$A_1: 2^1\times (2n - 1)$

$A_2: 2^2\times (2n - 1)$

$A_3: 2^3\times (2n - 1)$

$A_4: 2^4\times (2n - 1)$ etc.

So the n-th term of sequence $A_m$ must be $2^m\times (2n-1)$.

Now if $A_4$ contained 1000, that would mean that $16\times (2n-1) = 1000$ for some number n. But dividing 1000 by 16 gives a remainder, and 2n-1 is a whole number, so it can't be in $A_4$.

In fact, it can't be in $A_5$ either: we would need $32\times (2n-1) = 1000$, and if we can't divide 1000 by 16, then we can't divide it by 32 either!

So in fact it can't be in any sequence above $A_3$.

We also received similar excellent solutions from Andrea, Marjolaine, Misha, Shunya, Dmitris, Marcus and Alex from Highgate School, Krystof from Prague, and Aswaath from Garden School.

Holly, Harry and Simon also gave solutions to some of the later questions:

All the numbers from 1 to 63 can be found in some sequence, because you can always keep halving a number until it becomes odd (so is then in the first row). This means that any whole number can be found somewhere if you go on for long enough!

We do not think that a number can be repeated because each odd number only appears once. For example, for 44 to appear twice, 11 would have to appear twice, and it is only there once.

Aswaath also gave the same solutions to these questions, and listed the numbers from 1 to 63 and the sequences that they appeared in. It turned out that, out of the numbers between 1 and 63:

- 32 were in $A_0$
- 16 were in $A_1$
- 8 were in $A_2$
- 4 were in $A_3$
- 2 were in $A_4$
- and 1 was in $A_5$!

I wonder why...?

Aswaath also made the following comment:

The first sequence, $A_0$, will be the longest as it has all the positive odd numbers whereas the other sequences split up the positive evens between themselves.

Interesting! What do you think?

### Why do this problem?

This problem offers a really straightforward starting point for discussion of sequences that can lead on to generalisations, and perhaps for some students thinking about orders of infinity!

### Possible approach

$A_0 = 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29...$

$A_1 = 2, 6, 10, 14, 18, 22, 26, 30, 34, 38, 42...$

$A_2 = 4, 12, 20, 28, 36, 44, 52, 60...$

$A_3 = 8, 24, 40, 56, 72, 88, 104...$

$A_4 = 16, 48, 80, 112, 144...$

$A_5 = 32, 96, 160...$

$A_6 = 64...$

$A_7 = ...$

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Allow students some time to consider on their own or in pairs, noting down their thoughts before sharing them with the class.

Then pose the following question and allow students to continue working on their own or in pairs:**"Which sequences will contain the number 1000?**"

After a few minutes, hand out this scaffolding sheet

"When you've finished or can't make any further progress, look at the worksheet showing three approaches used by students working on this task."

"What might each student do next? Can you take each of their starting ideas and develop it into a solution?"

Here are some prompts that could be offered to students working on each approach if they get stuck:

For Alison's approach:

"What happens to the numbers as you go down the rows?"

"So what happens as you go up the rows?"

For Bernard's approach:

"Which numbers end in a 0 in row $A_2$?"

"Which numbers end in a 0 in row $A_3$?"

"Which of these sequences will hit 1000?"

For Charlie's approach:

"Can you find a similar method to Charlie's to describe the other rows?"

"Which descriptions include 1000?"

Select a few students to report back on how each approach was developed, and invite students to share their own alternative approaches.

**In a follow-up lesson,** return to the very first question "What do you notice?".

Invite students to phrase their noticings as questions and conjectures.

Here are some **key questions** that students might suggest, or which could be offered if students' ideas are not forthcoming (these appear at the bottom of the worksheet):

- How many of the numbers from 1 to 63 appear in the first sequence? The second sequence? ...

- Do all positive whole numbers appear in a sequence?
- Do any numbers appear more than once?
- Which sequence will be the longest?

- Given any number, how can you work out in which sequence it belongs?
- How can you describe the $n^{th}$ term in the sequence $A_0$? $A_1$? $A_2$? ... $A_m$?

### Possible support

Shifting Times Tables provides some preliminary work on sequences that may prepare students for tackling this task.

### Possible extension

Ask students to provide convincing proofs of their answers to two of the questions above:

- Do all positive whole numbers appear in a sequence?
- Do any numbers appear more than once?