# 2017 AIME I Problems/Problem 14

## Problem 14

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

so we have that $$256 = \log_a(\log_a(x)) = \frac{64}{3}\log_2\left(\frac{64}{3}\log_2(x)\right)$$ $$12 = \log_2\left(\frac{64}{3}\log_2(x)\right)$$ $$2^{12} = \frac{64}{3}\log_2(x)$$ $$192 = \log_2(x)$$ $$x = 2^{192}$$

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 Chinese Remainder Theorem, we get that $x\equiv \boxed{896}\bmod 1000$, finishing the solution.

## Alternate solution

If you've found $x$ but you don't know that much number theory.

Note $192 = 3 * 2^6$, so what we can do is take $2^3$ and keep squaring it (mod 1000). $$2^3 = 8$$ $$2^6 = 8*8 = 64$$ $$2^{12} = 64*64 \equiv 96\bmod 1000$$ $$2^{24} \equiv 96*96 \equiv 216\bmod 1000$$ $$2^{48} \equiv 216*216 \equiv 656\bmod 1000$$ $$2^{96} \equiv 656*656 \equiv 336\bmod 1000$$ $$2^{192} \equiv 336*336 \equiv \boxed{896}\bmod 1000$$

## Video Solution by mop 2024

~r00tsOfUnity

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