Prove that at least one integer in any set of ten consecutive integers is relatively prime to the others in the set.
Solution
There are 5 odd numbers in the set. At most 2 are multiples of 3, at most 1 is a multiple of 5 and at most 1 is a multiple of 7. So there is at least one odd number, k, that is not divisible by 3, 5 or 7. Now if k has a common factor with another member in the set, then that factor must divide their difference, which is at most 9. But the common factor cannot be divisible by 2, 3, 5 or 7, so it must be 1.
© John Scholes
jscholes@kalva.demon.co.uk
25 Jan 2002