AMC 2019 12A Problem 15 Diophantine Equation

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[et_pb_section fb_built="1" _builder_version="3.22.4" fb_built="1" _i="0" _address="0"][et_pb_row _builder_version="3.25" _i="0" _address="0.0"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||" _i="0" _address="0.0.0"][et_pb_text _builder_version="3.22.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" custom_padding="20px|20px|20px|20px" _i="0" _address="0.0.0.0"]

Understand the problem

[/et_pb_text][et_pb_text _builder_version="3.27" text_font="Raleway||||||||" background_color="#f4f4f4" box_shadow_style="preset2" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" _i="1" _address="0.0.0.1"]Positive real numbers $a$ and $b$ have the property that\[\sqrt{\log{a}} + \sqrt{\log{b}} + \log \sqrt{a} + \log \sqrt{b} = 100\]and all four terms on the left are positive integers, where log denotes the base 10 logarithm. What is $ab$?$\textbf{(A) }   10^{52}   \qquad        \textbf{(B) }   10^{100}   \qquad    \textbf{(C) }   10^{144}   \qquad   \textbf{(D) }  10^{164} \qquad  \textbf{(E) }   10^{200}$

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2019 AMC 12A Problems/Problem 15

[/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="3.27" hover_enabled="0" _i="1" _address="0.1.0.0.1" open="off"]logarithm, diophantine equation[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="3.27" hover_enabled="0" _i="2" _address="0.1.0.0.2" open="off"]Medium[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="3.27" hover_enabled="0" _i="3" _address="0.1.0.0.3" open="off"]Mathematical Circles (Russian Experience)[/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="3.27" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" _i="1" _address="0.1.0.1"]

Start with hints

[/et_pb_text][et_pb_tabs active_tab_background_color="#0c71c3" inactive_tab_background_color="#000000" _builder_version="3.27" tab_text_color="#ffffff" tab_font="||||||||" background_color="#ffffff" _i="2" _address="0.1.0.2"][et_pb_tab title="Hint 0" _builder_version="3.22.4" _i="0" _address="0.1.0.2.0"]Do you really need a hint? Try it first!

[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="3.27" hover_enabled="0" _i="1" _address="0.1.0.2.1"]Givenboth \( \sqrt {\log a} , \sqrt {\log b} \) are positive integers . \( \Rightarrow \) both \( \log a ,\log b \) are perfect squares .similarly , both \( \log {\sqrt a} , \log {\sqrt b} \) are positive integers.\( \Rightarrow \) both \( a ,b \) are perfect squares .[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.27" hover_enabled="0" _i="2" _address="0.1.0.2.2"]so $$  \log a = m^2 , \log b = n^2 $$ where \( m,n \in {Z^+} \)\( \Rightarrow a= 10^{m^2} , b= 10^{n^2} \) and as both \( a, b \) are perfect squares  \( \\ \Rightarrow 10^{m^2} ,10^{n^2}\) are both perfect squares i.e \( 10^{m^2} = p^2,10^{n^2} =q^2 \) , where \( p,q \in {Z^+} \) . \( \\ \Rightarrow \frac {m^2}{2} ,  \frac {n^2}{2} \) are integers \( \\ \Rightarrow 2|m^2 ,  2|n^2 \) \( \\ \Rightarrow 2|m,  2|n \) (as \(2\) is a prime number ) \( \\ \) so now \(  a= 10^{4x^2} , b= 10^{4y^2} \) can be put in the original equation , where  \( x,y \in {Z^+} \) . [/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.27" hover_enabled="0" _i="3" _address="0.1.0.2.3"] Now to get the solution from the derived equation i.e. \( 2x + 2y + 2x^2 +  2y^2 =100 \)  multiply both the sides by \( 2 \) and then add \( 2 \) in both sides to arrive at \( (2x+1)^2 + (2y+1)^2 = 202 \) .[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="3.27" hover_enabled="0" _i="4" _address="0.1.0.2.4"]Now use trial and error method to express \( 202 \) as a sum of two odd perfect squares .Finally the only way i.e. \(  9^2 + 11^2 = 202 \) . So without loss of generality it can be written that \( (2x+1) = 9 , (2y+1)= 11 \)So \( a= 10^{64} , b= 10^{100} \)  and \( ab = 10^{164} \)[/et_pb_tab][/et_pb_tabs][et_pb_text _builder_version="3.22.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" min_height="12px" custom_margin="50px||50px" custom_padding="20px|20px|20px|20px" _i="7" _address="0.1.0.7"]

Connected Program at Cheenta

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Similar Problems

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