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October 4, 2019

Number Theory - Croatia MO 2005 Problem 11.1

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Understand the problem

[/et_pb_text][et_pb_text _builder_version="3.27.4" text_font="Raleway||||||||" background_color="#f4f4f4" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" box_shadow_style="preset2" _i="1" _address="0.0.0.1"]Find all positive integer solutions of the equation $k!l! = k!+l!+m!.$

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Croatia MO 2005 [/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="3.29.2" hover_enabled="0" _i="1" _address="0.1.0.0.1" open="off"]Number Theory [/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="3.29.2" hover_enabled="0" _i="2" _address="0.1.0.0.2" open="off"]6/10 [/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="3.29.2" hover_enabled="0" _i="3" _address="0.1.0.0.3" open="on"]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" _i="1" _address="0.1.0.1"]

Start with hints

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[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="3.29.2" hover_enabled="0" _i="1" _address="0.1.0.2.1"]Observe that the LHS is sort of a quadratic and the RHS is sort of a linear type. Observe that the given equation is the same as (k!-1)(l! - 1) = m! + 1. Now observe that it implies that m > k,l. [/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.29.2" hover_enabled="0" _i="2" _address="0.1.0.2.2"]Given that m>l,k, Observe that the equation is symmetric in k and l. WLOG, assume l  \geq k. Now. there are two cases: 1. m > l = k. 2. m > l > k. [/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.29.2" hover_enabled="0" _i="3" _address="0.1.0.2.3"]1. m > l = k. Observe that it turns out that k! = 2 + (k+1)...m. Hence, k can be 0/1/2/3.  The only solution turns out to be (3,3,4) for (k,l,m). [/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="3.29.2" hover_enabled="0" _i="4" _address="0.1.0.2.4"]2. m > l > k. l! = 1 + (k+1)...l + (k+1)...l...m. Hence, it implies that l can only be = 1, but it doesn't give rise to any solution.  Hence, the only solution is (3,3,4). [/et_pb_tab][/et_pb_tabs][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" _i="3" _address="0.1.0.3"]

Watch video

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Connected Program at Cheenta

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Similar Problems

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