# Odds, Evens and More Evens

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:

"I have noticed that each number is double the number in the row above. I wonder if I can work out what would go in the rows above 1000?"

Bernard started by thinking:

"I have noticed that in $A_1$, the numbers which end in a 0 are 10, 30, 50... If I carry on going up in 20s I won't hit 1000, so I know 1000 isn't in $A_1$."

Charlie started by thinking:

"I have noticed that each number in $A_1$ is 2 more than a multiple of 4. I know 1000 is $250 \times 4$ so it can't be in $A_1$."

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 $A_0$? $A_1$? $A_2$? ... $A_m$?

## Comments

### Alison's Approach

NOTE: There are spoilers ahead.

So I decided to use Alison's idea. First, I proved the conjecture that each number is double the one above it using induction:

BASE CASE:
Clearly, the starting numbers in the series are double the ones above.
INDUCTIVE CASE:
Suppose the term before is double the one above it, like so:
$$k+n$$
$$2k+2n = j$$
Then we factor out the 2 and...
$$2(k+n) = j$$
Which was the desired result.

Combining the inductive and base cases, the theorem is proved.

The next step was realizing that I could half 1000 until I got an odd number, which are only in A0. It turns out I can half 1000 4 times, so 1000 is in A3.
But is 1000 anywhere else? We shall prove it cannot be using proof by contradiction.

If 1000 was in A3 more than one time, that would violate the fact that A3 is increasing.
That means that the "doubling steps" for 1000 are not equal to the "doubling steps" for 1000. But that's absurd!
Thus, that theorem is proved as well.

### I agree with some of your argument...

I agree with your argument why 1000 must be in A3, but I'm not sure about your use of induction. To use induction we would need to have some certain information about the sequences, but really you're being asked to deduce the pattern from examples.

### Maths

I believe that the answer is A3 because in A3 all the numbers are multiples of 8, but they miss out every other multiple of 8.
They miss out every even number of 8s.
e.g. there is 8, which is 1x8, but misses out 16, which is 2x8, then 24, which is 3x8, then misses out 32, which is an even number of 8s, and so on.
1000 divided by 8 gives you 125, which is an odd number, so 1000 will definitely appear in A3.

### Which sequences will contain the number 1000

Out of the sequences shown only A2 and A3 will contain the number 1000 because A0 will only have odd numbers = NO

In A1 you can't get to a 1000, because if you look at the first two numbers in the sequence with a last digit being 0 (10 and 30) there is a gap of 5 numbers between them in the sequence, and a gap of 20 between them, so if you add from 30 in 20's it will always be odd numbers:
30-50-70-90-110... = NO

And if you try and do it in 4's the closest you can get is 998-1002 = NO

A2 has a sequence of 4,12,20,28,36,44,52,60 and if you look at the two numbers that end with 0 there is a gap of 40 between them and 5 number between them in the sequence so if you go up from 60 in 40's: 60-100-140-180-220-260-300-340-380.....-800-840-880-920-960-1000 = YES

And the same with A3 which has a sequence of 8,24,40,56,72,88,104,120 so if you look again at the first two numbers ending with 0 (40 and 120) there is a 5 number gap in the sequence between them and an 80 gap between them so if you go from 120 up in 80's: 120-200-280-360-440.... 760-840-920-1000 = YES

If you carry on adding by either the difference between the two numbers ending in 0 from where there is a number ending in 0
eg: between 10-20 = 10 gap
20+10=30 +10=40......
Or adding by the number between every number in the pattern
eg: pattern 2,4,6,8,10= gap between each number is 2
so you add from 10 in 2s: 10+2=12 +2=14...
Or a mix of the two for each sequence you will see that no other sequence shown except A2 and A3 will contain 1000.

### Which sequences will contain the number 1000

I can follow your reasoning, but can you explain how you got from 380 to 800 going up in 40s?

### The A2 squence

How did you manage to jump from 380 to 800 by jumping in 40s?

(800-380)/40 = 420/40 = 10,5

### A2 sequence

Yes you are right A2 is actually NO because the closest it can get is 996 or 1004

### A2 doesn't contain 1000

The nth term of Am is double the nth term in Am-1.
If A2 had 1000, then A1 would have 500, and then A0 would have 250. Since A0 is made up of odd numbers, it can't contain 1000. You also miscounted. Here is what would be after 380.
420-460-500-540-580-620-660-700-740-780-820-860-900-940-980-1020.
A3 does contain 1000, as A2 would and does contain 500, which means that A1 contains 250, and A0 contains 125, which is odd. If you go any further than you would go into decimals,and no sequence contains a decimal.

### The answer

The way I got to the answer was by following Alison's train of thought.
If a sequence (An) is to contain 1000, the sequence above it (An-1) has to contain 500.
By that same logic, (An-2) has to contain 250.
And the one above that (An-3) has to contain 125.
Since A0 is the only sequence which contains odd numbers, and thus 125, A3 can be the only sequence which contains the number 1000.

### Which sequences will contain the number 1000 Charlie's way

If you look it at Charlie's way instead you can see that carrying on from where he started A2's sequence is the four times table just skipping out 1 number each time because it goes up in 8's so if you do 1000 divided by 4 =250 or divided by 8 just to make sure it's not a number that is skipped out it would =125 so A2=YES. A3's sequence is the 8 times table but skips out a numbers each time eg: 8-24(skips 16) because the sequence adds 16 each time so what you would do is 1000 divided by 8 = 125 = YES A4 is the 16 times table missing out a number each time so you would do 1000 divided by 16 which equals 62.5= NO A5 is the 32 times table missing out 1 each time so 1000 divided by 32 =31.25= NO A6 is the 64 times table missing out a number each time so do 1000 divided by 64 which equals 15.625=NO so out of the sequences shown only A2 and A3 have the number 1000

### Which sequences will contain the number 1000 Alison's way

If you do it Alison's way there is a bit of counting so if you think about what Alison said, each number is double the number in the row above.
To get to 1000 each sequence has to have the sequence above containing 500, which is less counting, and to make it even easier the row before that one has to have 250, and the one before that 125.
This way you can use the doubles as an advantage and do a lot less counting.

### A2 sequence

I was wrong on all three ways for sequence A2 as it does not have 1000 in its sequence as with Alison's way a1 doesn't reach 500 but 498 or 502 and with Bernard's way and Charlie's way it would equal 996 or 1004

### Persevering

Hi Sofia, well done for sticking at it with this problem even though you made a mistake initially, it's great to see you keeping going until you get the right answer!

### Math

I noticed that each sequence was a multiple of double the starting number take away the starting number so A2 is multiples of 8 (4x2) take away 4. So I had to find out if 1,004 would be divided 8 and it id not. So I found that 1,008 can be divided by 16 so A3 works.

### The Overall Method - I Call It 'Magic Fives'

After reading the answers which had been submitted, I managed to find a way to prove if 1000 would be in the sequence. As 5 will end in a 5 or a zero, you find the fifth term, carry adding that up until you reach 1000 or go above it. If you go above, it cannot be right, but if you hit 1000, then thats the correct sequence .
Remember - if the sequence is made up of odd numbers, it cannot be right.
I hope this helps in any way, shape or form!

### Equation for working out any term in any of the family members

My equation is:

$2(2^y)n-2^y\qquad$ where $n$ is the number of the term and $y$ is which number of sequence it is.

### Equation for working out any term in any of the family members

That looks like a very useful formula. So what does it tell you about which sequences contain the number 1000? Given any number, can you tell which sequence it will be in?