# Difference between revisions of "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.