2017 AIME I Problems/Problem 14

Problem

Let $a > 1$ and $x > 1$ satisfy $\log_a(\log_a(\log_a 2) + \log_a 24 - 128) = 128$ and $\log_a(\log_a x) = 256$ . Find the remainder when $x$ is divided by $1000$ .

Solution

The first condition implies

$a^{128} = \log_a\log_a 2 + \log_a 24 - 128$

$128+a^{128} = \log_a\log_a 2^{24}$

$a^{a^{128}a^{a^{128}}} = 2^{24}$

$\left(a^{a^{128}}\right)^{\left(a^{a^{128}}\right)} = 2^{24} = 8^8$

So $a^{a^{128}} = 8$ .

Putting each side to the power of $128$ :

$\left(a^{128}\right)^{\left(a^{128}\right)} = 8^{128} = 64^{64},$

so $a^{128} = 64 \implies a = 2^{\frac{3}{64}}$ . Specifically,

$\log_a(y) = \frac{\log_2(y)}{\log_2(a)} = \frac{64}{3}\log_2(y)$

so we have that

$256 = \log_a(\log_a(y)) = \frac{64}{3}\log_2\left(\frac{64}{3}\log_2(x)\right)$

$12 = \log_2\left(\frac{64}{3}\log_2(x)\right)$

$2^{12} = \frac{64}{3}\log_2(x)$

$192 = \log_2(x)$

$x = 2^{192}$

We only wish to find $x\bmod 1000$ . To do this, we note that $x\equiv 0\bmod 8$ and now, by the Chinese Remainder Theorem, wish only to find $x\bmod 125$ . By Euler's Theorem:

$2^{\phi(125)} = 2^{100} \equiv 1\bmod 125$

so

$2^{192} \equiv \frac{1}{2^8} \equiv \frac{1}{256} \equiv \frac{1}{6} \bmod 125$

so we only need to find the inverse of $6 \bmod 125$ . It is easy to realize that $6\cdot 21 = 126 \equiv 1\bmod 125$ , so

$x\equiv 21\bmod 125, x\equiv 0\bmod 8.$

Using CRT, we get that $x\equiv \boxed{896}\bmod 1000$ , finishing the solution.

2017 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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2017 AIME I Problems/Problem 14

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