#### You may also like Make a set of numbers that use all the digits from 1 to 9, once and once only. Add them up. The result is divisible by 9. Add each of the digits in the new number. What is their sum? Now try some other possibilities for yourself! ### Counting Factors

Is there an efficient way to work out how many factors a large number has? ### Repeaters

Choose any 3 digits and make a 6 digit number by repeating the 3 digits in the same order (e.g. 594594). Explain why whatever digits you choose the number will always be divisible by 7, 11 and 13.

# Differences

##### Age 11 to 14Challenge 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.