# 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 $ab$? $\textbf{(A) } 10^{52} \qquad \textbf{(B) } 10^{100} \qquad \textbf{(C) } 10^{144} \qquad \textbf{(D) } 10^{164} \qquad \textbf{(E) } 10^{200}$

# 2019 AMC 12A Problems/Problem 15

[/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"]Given both $\sqrt {\log a} , \sqrt {\log b}$ 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"]