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# IMO problem help? Watch

1. Find all positive integers n such that the set:
n, n+1, n+2, n+3, n+4, n+5
can be split into two disjoint subsets such that the products of the elements in these subsets is the same

I've said:
Since there are 6 numbers and 7 is prime they must be of the form
7n+1, 7n+2, ..., 7n+6
as 7 being a factor of any of the numbers would make it impossible

Which implies that the number of elements in each set must be 3, else the set with 4 or more would be much larger (6<7 and n=1 isn't a solution)

I've then listed the possible combinations that work (eliminating any where one would always be smaller than the other for positive n by comparing on a like for like magnitude)

I listed:
(7n+1)(7n+2)(7n+6) as one subset and the remaining terms as the other
along with
(7n+1)(7n+3)(7n+6)
(7n+1)(7n+4)(7n+5)
(7n+1)(7n+4)(7n+6)
(7n+1)(7n+5)(7n+6)
again, each making up one subset and the remaining terms making up the other.

Then expanded out both subsets, and in all but one of them every coefficient was larger in one of the subsets (and obviously all are positive, thus no solutions)

The one which wasn't was

(7n+1)(7n+4)(7n+6)=(7n+2)(7n+2)( 7n+5)

Which, when expanded and rearranged to (49n^2+21n-6=0) , gave irrational solutions.

The question implies that there is an n and I couldn't find one... where have I gone wrong?

(from the 12th IMO)
2. (Original post by MathMoFarah)

The question implies that there is an n and I couldn't find one... where have I gone wrong?
According to this there are no solutions.
3. (Original post by ghostwalker)
According to this there are no solutions.
Thanks!

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