AIME 2017 I · 第 14 题

Solution 1

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 Totient 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.

Solution 2 (Another way to find a)

$$\log_a(\log_a(\log_a 2) + \log_a 24 - 128) = 128$$ $$\implies \log_a(\log_a 2))+\log_a(24)=a^{128}+128$$ $$\implies \log_a(\log_a 2^{24})=a^{128}+128$$ $$\implies 2^{24}=a^{a^{(a^{128}+128)}}$$ Obviously letting $a=2^y$ will simplify a lot and to make the $a^{128}$ term simpler, let $a=2^{\frac{y}{128}}$. Then,

$$2^{24}=2^{\frac{y}{128} \cdot 2^{\frac{y}{128} \cdot (2^y+128)}}=2^{\frac{y}{128} \cdot 2^{y \cdot (2^{y-7}+1)}}$$ $$\implies 24=\frac{y}{128} \cdot 2^{y \cdot (2^{y-7}+1)}$$ $$\implies 3 \cdot 2^{10}=y \cdot 2^{y \cdot (2^{y-7}+1)}$$ Obviously, $y$ is $3$ times a power of $2$. Testing, we see $y=6$ satisfy the equation so $a=2^{\frac{3}{64}}$. Therefore, $x=2^{192} \equiv \boxed{896} \pmod{1000}$ ~Ddk001

Alternate solution 1

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).

$$2^3 = 8$$ $$2^6 = 8*8 = 64$$ $$2^{12} = 64*64 \equiv 96\bmod 1000$$ $$2^{24} \equiv 96*96 \equiv 216\bmod 1000$$ $$2^{48} \equiv 216*216 \equiv 656\bmod 1000$$ $$2^{96} \equiv 656*656 \equiv 336\bmod 1000$$ $$2^{192} \equiv 336*336 \equiv \boxed{896}\bmod 1000$$

Alternate solution 2

Another way to find $x \bmod 1000$ using modular arithmetic. In the same way as solution $1$, we can find that.

$$x\equiv 21\bmod 125, x\equiv 0\bmod 8.$$ $$x = 8m = 125n+21$$ For some positive integers $m$ and $n$. Taking the equation mod $8$ gives

$$5n+5 \equiv 0\bmod 8$$ $$n \equiv 7\bmod 8$$ $$n = 8k-1$$ For some positive integer $k$. Plug this back into the original equation.

$$8m = 125(8k-1)+21$$ $$8m = 1000k-104$$ $$x = 8m = 1000k - 104$$ $$x \equiv -104 \equiv 896\bmod 1000$$ $$x \equiv 896\bmod 1000$$ ~sdfgfjh

Video Solution by mop 2024

https://youtu.be/E-7YQ9ND5Ms

~r00tsOfUnity

Video Solution 2 by SpreadtheMathLove

https://www.youtube.com/watch?v=ZVPhvjlBvWY