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Combinatorics Difficulty 7.5 National olympiad, round 2 Prove it Argentina

A company has nn employees. It is known that every employee works at least one of the 7 days of the week, except for an employee that does not work any of the 7 days. In addition, for every pair of these nn employees, there are at least 3 days of the week such that one of the two employees works that day but the other does not (it is not necessarily the same employee who works those days). Find the maximum possible value of nn.

Solution

We will show that the maximum possible number of employees is n=16n = 16.

First, we prove that n16n \le 16. We represent each possible weekly schedule for an employee with a 7-tuple of 0's and 1's whose coordinates correspond to the days of the week and where 1 stands for 'working day' and 0 stands for 'non working day'.

The condition of the problem states that, for every pair of employees, their weekly schedules differ in at least 3 days, that is, the corresponding tuples differ in at least 3 coordinates.

Let v1,,vnv_1, \ldots, v_n be the tuples representing the weekly schedules of the nn employees. For each viv_i, there are 7 tuples differing in exactly one coordinate with viv_i; we call them neighbors of viv_i. Note that, if iji \ne j, vjv_j is not a neighbor of viv_i, since they differ in at least 3 coordinates; in addition, viv_i and vjv_j do not have common neighbors, since if there exists vv differing in exactly one coordinate with viv_i and exactly one coordinate with vjv_j, then viv_i and vjv_j differ in at most 2 coordinates, contradicting the assumption.

Then, considering v1,,vnv_1, \ldots, v_n and all their neighbors, we have n+7n=8nn + 7n = 8n different tuples. Since there are 27=1282^7 = 128 possible weekly schedules, we deduce that 8n1288n \le 128, which implies that n16n \le 16.

Finally, the following is an example with 16 employees satisfying the required conditions:

v1=(0,0,0,0,0,0,0),v2=(1,1,1,0,0,0,0),v3=(1,0,0,1,1,0,0),v4=(1,0,0,0,0,1,0),v5=(0,1,0,1,0,0,1),v6=(0,1,0,0,1,1,0),v7=(0,0,1,1,0,1,0),v8=(0,0,1,0,1,0,1),v9=(1,1,1,1,1,1,1),v10=(0,0,0,1,1,1,1),v11=(0,1,1,0,0,1,1),v12=(0,1,1,1,1,0,0),v13=(1,0,1,0,1,1,0),v14=(1,0,1,1,0,0,1),v15=(1,1,0,0,1,0,1),v16=(1,1,0,1,0,1,0). \begin{align*} v_1 &= (0, 0, 0, 0, 0, 0, 0), & v_2 &= (1, 1, 1, 0, 0, 0, 0), & v_3 &= (1, 0, 0, 1, 1, 0, 0), & v_4 &= (1, 0, 0, 0, 0, 1, 0), \\ v_5 &= (0, 1, 0, 1, 0, 0, 1), & v_6 &= (0, 1, 0, 0, 1, 1, 0), & v_7 &= (0, 0, 1, 1, 0, 1, 0), & v_8 &= (0, 0, 1, 0, 1, 0, 1), \\ v_9 &= (1, 1, 1, 1, 1, 1, 1), & v_{10} &= (0, 0, 0, 1, 1, 1, 1), & v_{11} &= (0, 1, 1, 0, 0, 1, 1), & v_{12} &= (0, 1, 1, 1, 1, 0, 0), \\ v_{13} &= (1, 0, 1, 0, 1, 1, 0), & v_{14} &= (1, 0, 1, 1, 0, 0, 1), & v_{15} &= (1, 1, 0, 0, 1, 0, 1), & v_{16} &= (1, 1, 0, 1, 0, 1, 0). \end{align*}

We denote d(vi,vj)d(v_i, v_j) the number of coordinates in which viv_i and vjv_j differ. Note that:

* d(v1,v3)=3d(v_1, v_3) = 3, for 2j82 \le j \le 8, d(v1,v9)=7d(v_1, v_9) = 7, and d(v1,vj)=4d(v_1, v_j) = 4, for 10j1610 \le j \le 16;
* for 2i82 \le i \le 8, d(vi,vj)=4d(v_i, v_j) = 4, for 2j92 \le j \le 9, jij \ne i, d(vi,vj)=4d(v_i, v_j) = 4 for 10j1610 \le j \le 16, ji+8j \ne i+8, and d(vi,vi+8)=7d(v_i, v_{i+8}) = 7;
* d(v9,vj)=3d(v_9, v_j) = 3, for 10j1610 \le j \le 16, and d(vi,vj)=4d(v_i, v_j) = 4 for 10i<j1610 \le i < j \le 16.

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