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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.

• August 13, 2011 at 8:09 am

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

2. July 22, 2011 at 11:31 pm

1 5 7 11
4 pairs

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

and the nA is 4.

5. 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. 🙂

• February 25, 2013 at 7:27 pm

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