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Difference between revisions of "2017 AIME I Problems/Problem 14"
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<cmath>2^3 = 8</cmath>
 
<cmath>2^3 = 8</cmath>
 
<cmath>2^6 = 8*8 = 64</cmath>
 
<cmath>2^6 = 8*8 = 64</cmath>
−
<cmath>2^{12} = 64*64 = 96 (mod 1000)</cmath>
+
<cmath>2^{12} = 64*64 \equiv 96\bmod 1000</cmath>
−
<cmath>2^{24} = 96*96 = 216 (mod 1000)</cmath>
+
<cmath>2^{24} \equiv 96*96 \equiv 216\bmod 1000</cmath>
−
<cmath>2^{48} = 216*216 = 656 (mod 1000)</cmath>
+
<cmath>2^{48} \equiv 216*216 \equiv 656\bmod 1000</cmath>
−
<cmath>2^{96} = 656*656 = 336 (mod 1000)</cmath>
+
<cmath>2^{96} \equiv 656*656 \equiv 336\bmod 1000</cmath>
−
<cmath>2^{192} = 336*336 = \boxed{896} (mod 1000)</cmath>
+
<cmath>2^{192} \equiv 336*336 \equiv \boxed{896}\bmod 1000</cmath>
  
 
== See also ==
 
== See also ==

Revision as of 15:04, 2 February 2023

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).

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

See also

2017 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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