2019 AMC 12A Problems/Problem 15

Problem

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 $ab$ ?

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

Solution

Since all four terms on the left are positive integers, from $\sqrt{\log{a}}$ , we know that both $\log{a}$ has to be perfect square and $a$ has to be a power of ten. The same applies to $b$ for the same reason. Setting $a$ and $b$ to $10^x$ and $10^y$ , where $x$ and $y$ are the perfect squares, $ab = 10^{x+y}$ . By listing all the perfect squares up to $14^2$ (as $15^2$ is larger than the largest possible sum of $x$ and $y$ of $200$ from answer choice $E$ ), two of those perfect squares must add up to one of the possible sums of $x$ and $y$ given from the answer choices ( $52$ , $100$ , $144$ , $164$ , or $200$ ).

Only a couple possible sums are seen: $16+36=52$ , $36+64=100$ , $64+100=164$ , $100+100=200$ , and $4+196=200$ . By testing each of these (by seeing whether $\sqrt{x} + \sqrt{b} + \frac{x}{2} + \frac{y}{2} = 100$ ), only the pair $x = 64$ and $y=100$ work. Therefore, $a$ and $b$ are $10^{64}$ and $10^{100}$ , and our answer is $\boxed{\textbf{(D) } 100^{164}}$ .

2019 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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2019 AMC 12A Problems/Problem 15

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