# Differences

* Differences printable worksheet*

Choose any three whole numbers, find the differences between them all, and find the product of the differences.

For example, if your three whole numbers are $7$, $4$ and $12$, the differences are:

$12 - 7 = 5$

$12 - 4 = 8$

$7 - 4 = 3$

The product of the differences is

$3\times 5\times 8 = 120$

Try a few examples.

**What do you notice?Can you explain what you've noticed?**

Now choose any four whole numbers, find the differences between them all, and find the product of the differences.

For example, if your four whole numbers are $7$, $4$, $12$ and $6$, the differences are:

$7-4 = 3$

$12-7 = 5$

$7-6 = 1$

$12-4 = 8$

$6-4 = 2$

$12-6 = 6$

The product of the differences is $3\times 5\times 1\times 8\times 2\times 6 = 1440$

Try a few examples.

**What do you notice?Can you explain what you've noticed?**

Pick three test values - calculate differences (why are only three listed?) - and then the product. Can you find an exception? The question is "Why not?".

The second half of the question is about multiples of three. Can you 'categorise' numbers in terms of their relationship to mulltiples of three and how does this help?

Have you seen the problem Take Three from Five ?

There were many solutions to the first part of this question based on odd and even numbers.

Alexander from Shevah-Mofet School,
Israel and also the Key Stage 3 Maths Club from Strabane Grammar
School proved that, for any distinct whole numbers
*a* ,
*b* and
*c* , the
expression

( *a*
- *b*
)( *a*
- *c*
)( *b*
- *c*
) is divisible by 2.

At least two of the distinct whole numbers *a* ,
*b* or *c* must have the same remainder when divided
by 2 (because there are only two possibilities), therefore their
difference will be an even number and ( *a* - *b* )(
*a* - *c* )( *b* - *c* ) must be
even

Alexander generalised this solution:

With four distinct whole numbers the same idea is used to prove
that( a - b )( a - c )( a - d )(
b - c )( b - d )( c - d) is divisible by 3. At least two of
the numbers *a* , *b* , *c* or *d* have
the same remainder when divided by 3; therefore one of the
differences divides by 3 and so the entire thing divides by 3.

Taking *n* distinct whole numbers, the product of the
differences of the numbers taken in pairs is divisible by (
*n* -1) because at least two of the numbers have the same
remainder when divided by ( *n* -1).

The Maths Club at Wilson's School sent in this very clearly explained proof:

If we have a set of n+1 integers, then by the pigeonhole principle, two of these integers must have the same value modulo n. Therefore the difference between these 2 values will be congruent to 0 modulo n, and so will be a multiple of n. Thus the product of all the remainders will be a multiple of n.

Also because in a set of n+1 integers you always have 2 values that the same modulo n-1, n-2, n-3 ... , 2 , 1, we can say that the product of the differences of n+1 numbers will be divisible by n!

Ruth from The Manchester High School for Girls proved that the product of the differences of four numbers is not necessarily a multiple of 5. For example the 4 numbers 1, 2, 3 and 4. The product of the differences is (2-1)(3-2)(3-1)(4-3)(4-2)(4-1)=1*1*2*1*2*3=12 which is not a multiple of 5.

For further discussion of this problem see Modulus Arithmetic and a Solution to Differences by Peter Zimmerman (Mill Hill County High School, London). Peter has given a very good account of modulus arithmetic which not only provides a solution to this problem but will also help you to understand a method which can be applied to many other problems.

### Why do this problem?

At the heart of this problem is using a representation which helps to simplify the situation. For example, in the first part, using the fact that any number is either odd or even reveals the structure. The context and argument can be based on modulo arithmetic. Working from specific cases in order to generalise is a mathematical technique that can also be highlighted.

Possible approach

- Can they find three numbers where this is not the case?
- Why not?

- For the product not to be even what can you say about the differences?
- What does this mean about the original three numbers?

Now move on to the second part of the problem. Some help with describing the three types of number related to multiples of three might be needed (see the notes to the problem Take Three from Five ).

### Key questions

- Why are there only three and six differences in the lists? (because $(a-b)$ is numerically equal to $(b-a) etc$)
- How can we describe all numbers in terms of muliples of $3$, or $4$ or $5$ ...?
- For the product not to be a multiple of three what can you say about the differences?

### Possible support

Make 37 might be good to try first.

### Possible extension

How many integers do you need to ensure that the product of all the differences is divisible by $5$?

Some students may go on to investigate this context more thoroughly, including posing and pursuing their own questions. For example: What about divisibility by $4$ and $6$, and then more generally?

Odd Stones and Take Three from Five might provide suitable follow-up problems.