Challenge Level

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?