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IMO 2011 Problem 1


Given any set A = \{a_1, a_2, a_3, a_4\} of four distinct positive integers, we denote the sum a_1 +a_2 +a_3 +a_4 by s_A. Let n_A denote the number of pairs (i, j) with 1\leq i < j\leq 4 for which a_i +a_j divides s_A. Find all sets A of four distinct positive integers which achieve the largest possible value of n_A.

IMO 2011 Problem 1

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  1. July 20, 2011 at 5:31 pm

    If the four numbers are equal then n_A = 6. Now since this is the maximum possible, let’s look for the general form of these sets.

    Then: s_A \equiv 0 \pmod{a_1 + a_2} and s_A \equiv 0 \pmod{a_3 + a_4}
    which is equivalent to: a_3 + a_4 \equiv 0 \pmod{a_1 + a_2} and a_1 + a_2 \equiv 0 \pmod{a_3 + a_4}

    \Rightarrow a_1 + a_2 = a_3 + a_4

    Similarly we get, a_1 + a_3 = a_2 + a_4 and a_1 + a_4 + a_2 + a_3

    Solving this we get a_1 = a_2 = a_3 = a_4, which satisfies n_A = 6

    • July 21, 2011 at 3:37 pm

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

    • cvghvb
      August 13, 2011 at 8:09 am

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

  2. kris
    July 22, 2011 at 11:31 pm

    1 5 7 11
    4 pairs

  3. mariucell
    July 24, 2011 at 10:36 am

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

  4. mariucell
    July 24, 2011 at 10:36 am

    and the nA is 4.

  5. Ernesto H. Escobedo
    July 25, 2011 at 9:15 pm

    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.

    • July 25, 2011 at 9:28 pm

      Thank you for your solution. 🙂

      • Andy
        February 25, 2013 at 7:27 pm

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

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