AIME 2017 I – Problem 14

Solution: Let $a=2^n$, we have $$log_{2^n}(\log_{2^n}(\log_{2^n}2)+\log_{2^n}24-128)=128$$

$$log_{2^n}(\log_{2^n}2)+\log_{2^n}24-128=2^{128n}$$

$$log_{2^n}(\log_{2^n}2^{24})=2^{128n}+128$$

$$log_{2^n}(\dfrac{24}{n})=2^{128n}+128$$

$$\dfrac{1}{n}log_2(\dfrac{24}{n})=2^{128n}+128$$

$$log_2(\dfrac{24}{n})=n(2^{128n}+128)$$

Let $n=3\cdot 2^m$, we have $$3-m=3\cdot 2^m(2^{384\cdot 2^m}+128)$$

Check the range of $m$, we found $-7<m<-5$ and $m=-6$ is a solution. Therefore $n=3\cdot 2^{-6}=\dfrac{3}{64}$, and $a=2^{\frac{3}{64}}$. As $log_ax=a^{256}$, we have

$$log_{2^{\frac{3}{64}}}x=2^{256\cdot{\frac{3}{64}}}=2^{12}$$

$$\dfrac{64}{3}log_{2}x=2^{12}$$

$$log_{2}x=192$$

$$x=2^{192}$$

$$x|1000=2^{192}|1000=4\cdot 2^{190}|1000=4\cdot 1024^{19}|1000=4\cdot 24^{19}|1000$$

$$=96\cdot 24^{18}|1000=96\cdot 13824^6|1000=96\cdot 176^6|1000=96\cdot 30976^3|1000$$

$$=-96\cdot 24^3|1000=-96\cdot 13824|1000=96\cdot 176|1000=16896|1000=\boxed{896}$$