# Different by One

*If you have not used the Cuisenaire Environment before, you might find it helpful to look at the instructions and video before trying this problem.*

Choose two different colours of rod, and make a line of each colour.

Can you make your two lines differ by the length of one white rod?

For example, here are three greens and two yellows. The length of the greens is one less than the length of the yellows.

They could also be arranged like this:

Try some different pairs of rods and look for arrangements where the lengths differ by one.

Here are some questions you might like to explore:

If you start with two rods that differ in length by 2, when is it possible to make two lines that differ by 1?

How many of each rod will you need?

What if you could use rods longer than 10, for example 23 and 25? Or 99 and 101? Or...

Once you've found an arrangement where the two lines differ by one, is there a quick way to find other arrangements with the same colour rods that also differ by one?

What if you start with two rods that differ in length by 3? By 4?

In the original example, the lines were made from rods that differed in length by 2:

light green rods (3) and yellow rods (5).

You could start by creating lines made from

yellow (5) and black (7)

black (7) and blue (9)...

Which pairs can be used to create lines that differ by 1?

This was a challening problem! Well done to everyone who shared their thoughts and solutions with our team.

Students at Long Field Academy began by exploring this problem using the Cuisenaire environment. Theo explained how to find rods that differ in length by one:

To create a difference of one white rod it is all about times tables. If you were to have three rods of three, which would be 3x3=9, and then if you had two rods of four which is 2x4=8 which creates one rod difference. Another way you could do it is 7x3 and 4x5 which would be 21 and 20.

Caleb from Kilvington Grammar in Australia used rods that differ in length by 2, and explained when it is impossible to get a difference of 1:

To solve this problem, we start with two terms: $x$ and $(x-2).$ The first thing I figured out is that $x$ cannot be even, because then both terms would be even, and anything multiplied by an even number is an even number, therefore not giving us a difference of $1,$ which would only appear in an even and an odd number. Also, using the equation below, you wouldn't multiply the number by a whole number. I don't like my solution, because I just used trial and error, but here it is:

$x\times\frac{x-1}2 - (x-2)\times\frac{x+1}2=1$ [*that is, $\frac{x-1}2$ rods of length $x$ and $\frac{x+1}2$ rods of length $(x-2)$ have a difference of $1$*]

Yinuo from HD Ningbo School in China had the same formula, but Yinuo called the lengths $l_1$ and $l_2$ instead of $x$ and $(x-2)$. Yinuo proved that the formula really does give a difference of $1$:

Mahdi from Mahatma Gandhi International School proved that you can never make a difference of $1$ when the rods differ by $2$ and both have even length:

*(Mahdi is talking about the difference multiplied by 2)*

Mahdi then considered the case where the rods have odd lengths, and found an alternative proof to the ones shown above. Click to see Mahdi's work.

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note: the denominator should say "Number of numbers", or something similar, not "Difference"

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note: the denominator should say "Number of numbers", or something similar, not "Difference"

Yinuo then thought about rods with lengths that differ by 3 and 4. This is Yinuo's work:

i) Two rods that are multiples of $3$

No arrangement

ii) $l_1-l_2=3,$ $l_1$ and $l_2$ are both greater than multiples of $3$ by $1,$ e.g. $1, 4$ and $4,7$*In the equation below, Yinuo writes $3n+1$ in brackets to add information about $l_1$ and $l_2$. It does not mean that $l_1$ and $l_2$ are multiplied by $3n+1$.*

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*Click to see a proof.*

$\begin{split}\frac{l_1-1}3l_2 - \frac{l_2-1}{3}l_1 &= \tfrac{1}{3}\times\left((l_1-1)l_2 - (l_2-1)l_1)\right)\\

& =\tfrac{1}{3}\left(l_1l_2-l_2 - l_1l_2+l_1\right)\\

&=\tfrac{1}{3}\left(l_1-l_2\right)\\

&=1\end{split}$

iii) $l_1$ and $l_2$ are both less than multiples of $3$ by $1,$ e.g. $2, 5$ and $5,8$ *(equivalently these are $2$ more than a different multiple of $3$)*

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*Click to see a proof.*

$\begin{split}\frac{l_2+1}{3}l_1 - \frac{l_1+1}3l_2 &= \tfrac{1}{3}\times\left((l_2+1)l_1 - (l_1+1)l_2)\right)\\

& =\tfrac{1}{3}\left(l_1l_2+l_1 - l_1l_2-l_2\right)\\

&=\tfrac{1}{3}\left(l_1-l_2\right)\\

&=1\end{split}$

i) Two rods have even lengths

No arrangements

ii) $l_1-l_2=3,$ $l_1$ and $l_2$ are both greater than multiples of $4$ by $1,$ e.g. $1, 5$

iii) $l_1-l_2=3,$ $l_1$ and $l_2$ are both smaller than multiples of $4$ by $1,$ e.g. $3, 7$

Yinuo says that possible arrangements exist "periodically". Mahdi showed how to use one arrangement which gives a difference of $1$ to find other arrangements which also give a difference of $1$:

Let the two rods be of difference $d.$ Thus let the lengths be $k, k+d.$

Let there be $x$ number of $k$ rods and $y$ number of $d$ rods which give a difference of $1.$

__To get infinite pairs:__

Find the lowest common multiple (LCM) of $k$ and $k+d$. If they are mutually prime, the LCM is $k(k+d)$ *(if they have no common factors, they are said to be coprime, which is sufficient here)*

Adjust $n$ to any value and get infinite pairs.

$\left(x + n\times\frac{\text{LCM}}k, y + n\times\frac{text{LCM}}k\right)$

If they are prime *(or coprime)*:

$\left(x + n(k + d), y + nk\right)$

### Why do this problem?

This problem presents a context in which students can work systematically, notice surprising patterns, and make generalisations that they can explain visually, numerically or algebraically.### Possible approach

Students could use Cuisenaire rods to work on this challenge, or the online Cuisenaire environment, or squared paper if rods are not available. Introduce the problem: invite students to choose two different colours of rod and make a line of each colour so that the lengths differ by the length of one white rod. You might like to show them the examples using yellow and green:Give students some time to explore, and then challenge students to look for general patterns in their results. One very fruitful line of enquiry, which leads to generalisations that might involve triangular numbers, is suggested in the problem:

- If you start with two rods that differ in length by 2, when is it possible to make two lines that differ by 1?
- How many of each rod will you need?

Some students might use algebra to justify the patterns they notice.

Another line of enquiry might be to consider how to generate other arrangements from a particular example.

For example:

7 greens is longer by 1 than 4 yellows

12 greens is longer by 1 than 7 yellows

### Key questions

Using red (2) and pink (4), you can never make a difference of 1 - why not?How can you tell whether it will be possible to make a difference of 1 for a particular pair of rods?

If you record your results in a systematic way, what patterns can you see? Can you prove that they will continue?

### Possible support

Encourage students to start by investigatingyellow (5) and black (7)

dark green (6) and brown (8)

black (7) and blue (9)