Is there an efficient way to work out how many factors a large number has?
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.
Find the number which has 8 divisors, such that the product of the
divisors is 331776.
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
c , the
) 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
Alexander generalised this
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
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
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