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## IMO 2011 Problem 1

Given any set of four distinct positive integers, we denote the sum by . Let denote the number of pairs with for which divides . Find all sets of four distinct positive integers which achieve the largest possible value of .

*IMO 2011 Problem 1*

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Categories: Uncategorized
Combinatorics, IMO

If the four numbers are equal then . Now since this is the maximum possible, let’s look for the general form of these sets.

Then: and

which is equivalent to: and

Similarly we get, and

Solving this we get , which satisfies

The numbers are distinct. Your solution is not valid. I’m sorry.

it says distinct integers !!! they can’t be equal !!!!! 🙂

1 5 7 11

4 pairs

I have found this solutions: (c, 5c, 7c, 11c) and (d,11d, 19d, 29d), with c and d positive integers.

and the nA is 4.

Consider the 6 sums with i>j. If we choose 4 of them to be divisors of Sa, we will always choose some ai such that the sums ai+aj, ai+ak, ai+an (with (i,j,k,n) = (1,2,3,4) in some order) are chosen. Let:

(ai+aj)= Sa/k1

(ai+ak)=Sa/k2

(ai+an)=Sa/k3

Summing up, we get Sa + 2ai = Sa(1/k1 + 1/k2 + 1/k3) hence:

1/k1 +1/k2 +1/k3 – 1 = 2ai/Sa

None of the k can be 1, since the elements are positive. Then the only possible triplets of k that provide 1/k1 +1/k2 +1/k3>1 are k1=2, k2=3, k3=4,5. If k3=4, ai=Sa/24, ak=11Sa/24, aj=7Sa/24, an=5Sa/24, which gives us na=4. If k3=5, ai= Sa/60, aj=29Sa/60, ak=19Sa/60, an=11Sa/60, which also gives us na=4.

Thank you for your solution. 🙂

Best solution generated for the problem I ever saw. Respect!