2019 AMC 12A Problem 15

Attempt Problem 15 of the 2019 AMC 12A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2019 AMC 12A solutions, or check the answer key.

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15.

Positive real numbers

a

and

b

have the property that

\begin{aligned} &\sqrt{\log a} + \sqrt{\log b} \\ &\quad {}+ \log \sqrt{a} + \log \sqrt{b} = 100 \end{aligned}

and all four terms on the left are positive integers, where

\log

denotes the base

10

logarithm. What is

ab?

10^{52}

10^{100}

10^{144}

10^{164}

10^{200}

Answer: D

Concepts:logarithm Diophantine Equation casework

Difficulty rating: 1730

Small hint:

Let

\sqrt{\log a} = p

and

\sqrt{\log b} = q;

then

\log\sqrt{a} = \tfrac{p^2}{2}

must be an integer

Big hint:

With

p = 2m,

q = 2n,

the equation becomes

m(m+1) + n(n+1) = 50

Solution:

Let

\sqrt{\log a} = p

and

\sqrt{\log b} = q,

so

\log a = p^2

and

\log\sqrt{a} = \dfrac{p^2}{2}.

For this to be an integer,

p

is even; likewise

q.

Writing

p = 2m,

q = 2n,

the equation

p + q + \dfrac{p^2}{2} + \dfrac{q^2}{2} = 100

becomes

m(m+1) + n(n+1) = 50.

The only solution is

\{m, n\} = \{4, 5\},

giving

\log(ab) = p^2 + q^2

= 4(16 + 25) = 164.

Therefore

ab = 10^{164}.

Thus, the correct answer is D.

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