Challenge Level

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.