Expert: Ahmed Salami - 11/3/2008

Question
Find, with proof, the largest positive integer k with the following property:
There exists a positive number N such that N is divisible by all but three of the
integers 1, 2, 3, . . . , k, and furthermore those three integers (that don’t divide N)
are consecutive


Answer
Hi Socra,
For starters, N is never divisible by N-1 except when N = 2 and that clearly doesnt satisfy us because 2 doesnt have up to three positive integers below it.
That means that N-1 is one of our three numbers and quite rightly the largest of them, the other numbers are then N-2 and N-3. Considering this we then agree that N-4 is a factor of N.
Take N = p(N-4) where p is an integer constant.
N = pN - 4p
N - pN = -4p
N(1-p) = -4p
N = -4p/(1-p)
N = 4p/(p-1)
N is an integer and so 4p/(p-1) is an integer and we conclude that 4p is divisible by p-1. But p-1 is also a factor of 4(p-1). It'll therefore be a factor of their difference i.e
4p - 4(p-1) = 4
4 is divisible by p-1, but 4 is only divisible by 1,2 and 4. So,
p-1 = 1
p = 2
OR
p-1 = 2
p = 3
OR
p-1 = 4
p = 5
For p = 2, N = 8/1 = 8 and N-4 = 4. But this doesnt satisfy our condition because apart from 5,6,7, we still have 3 which doesnt divide 8.
For p = 3, N = 12/2 = 6 and N-4 = 2. This also fails us because 3, one of our three numbers that shouldnt divide 6, actually divides 6.
For p = 5, N = 20/4 = 5 and N-4 = 1
This is the solution that satisfies us because 2,3,4 dont divide 5 but 1 does and so the largest integer is 5.

Regards.