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October 22, 2019

RMO 2019 Problem 6 Solution

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Understand the problem

[/et_pb_text][et_pb_text _builder_version="4.0" text_font="Raleway||||||||" background_color="#f4f4f4" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" box_shadow_style="preset2"]Suppose 91 distinct positive integers greater than 1 are given such that there are at least 456 pairs among
them which are relatively prime. Show that one can find four integers a, b, c, d among them such that
gcd(a, b) = gcd(b, c) = gcd(c, d) = gcd(d, a) = 1.

[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.25"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="4.0" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px"][et_pb_accordion_item title="Source of the problem" open="on" _builder_version="4.0"]Regional Math Olympiad, 2019 Problem 6

[/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="4.0" open="off"]Combinatorics

[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="4.0" open="off"]

8/10[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="3.29.2" open="off"]Challenges and Thrills in Pre College Mathematics[/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

Start with hints

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[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="4.0"]The clue: Consider the numbers as points in a graph, and some edges between the points based on their gcd. Let us consider a graph G with 91 vertices (91 distinct numbers) and connect each pair of these vertices. which corresponds to co-prime pairs implies at least 456 edges i.e. e \( \geq\) 456. Here e = number of edges. Now getting four number a,b, c, d such that gcd (a, b) = gcd (b, c) = gcd(c,d) = gcd (d,a) = 1 implies existence at a cycle of length 4.  [/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="4.0"]We will follow the contradiction method. Let us assume there is no cycle of length '4'. Our aim is to find a bound on e, which contradicts the 456 bound on e. Let vertex set of G be V =  {\(v_1, v_2,...,v_{91} \)}, let degree of \( v_i = d_i\). Now, observe that the number of vertex pairs {\( v_k, v_l\)}, adjacent to \(v_i\) is \({d_i}\choose{2}\). Now observe that a vertex pair {\( v_k, v_l\)} cannot be common to two vertices say \(v_i\) and \(v_j\) both, because if so, \( v_i, v_k, v_j, v_l, v_i\) will form a cycle.[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="4.0"]

So, the sum of all such vertex pairs corresponding to a vertex over all the vertex set \( \leq\) the total number of vertex pairs. \( \sum_{i=1}^{91} ({d_i}\choose{2}) \leq {{91}\choose{2}} \rightarrow \) \( \sum_{i=1}^{91} ({d_i}^2 - d_i)\leq 91.90  \rightarrow \)  \( \sum_{i=1}^{91} {d_i}^2 \leq 2e + {{91}\choose{2}}\) -> (1)  as \( \sum_{i=1}^{91} d_i = 2e \).

We will try to find a lower bound on  \( \sum_{i=1}^{91} {d_i}^2 \) w.r.t e.

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By RMS - AM inequality, \( \sum_{i=1}^{91} {d_i}^2 \geq  \frac{( \sum_{i=1}^{91} {d_i})^2 }{91} = \frac{4e^2}{91} \) -> (2) Using (1) and (2), we get \( \frac{2e^2}{91} - e \leq 91.45 \) From here, by solving we get \( e \leq 91x5 =455\). This gives the contradiction.

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Watch video

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Connected Program at Cheenta

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Similar Problems

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