Difference between revisions of "2017 AIME I Problems/Problem 14"

Latest revision as of 14:29, 8 September 2024

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 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

@@ Line 53: / Line 53: @@
 ==Solution 2 (Another way to find a)==
-<math>\log_a(\log_a(\log_a 2) + \log_a 24 - 128) = 128</math>
+<cmath>\log_a(\log_a(\log_a 2) + \log_a 24 - 128) = 128</cmath>
-<math>\implies \log_a(\log_a 2))+\log_a(24)=a^{128}+128</math>
+<cmath>\implies \log_a(\log_a 2))+\log_a(24)=a^{128}+128</cmath>
-<math>\implies \log_a(\log_a 2^{24})=a^{128}+128</math>
+<cmath>\implies \log_a(\log_a 2^{24})=a^{128}+128</cmath>
-<math>\implies 2^{24}=a^{a^{(a^{128}+128)}}</math>
+<cmath>\implies 2^{24}=a^{a^{(a^{128}+128)}}</cmath>
 Obviously letting <math>a=2^y</math> will simplify a lot and to make the <math>a^{128}</math> term simpler, let <math>a=2^{\frac{y}{128}}</math>. Then,
-<math>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)}}</math>
+<cmath>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)}}</cmath>
-<math>\implies 24=\frac{y}{128} \cdot 2^{y \cdot (2^{y-7}+1)}</math>
+<cmath>\implies 24=\frac{y}{128} \cdot 2^{y \cdot (2^{y-7}+1)}</cmath>
-<math>\implies 3 \cdot 2^{10}=y \cdot 2^{y \cdot (2^{y-7}+1)}</math>
+<cmath>\implies 3 \cdot 2^{10}=y \cdot 2^{y \cdot (2^{y-7}+1)}</cmath>
-Obviously, y is <math>3</math> times a power of <math>2</math>. (It just makes sense.) Testing, we see <math>y=6</math> satisfy the equation so <math>a=2^{\frac{3}{64}}</math>. Therefore, <math>x=2^{192} \equiv \boxed{896} \pmod{1000}</math> ~[[Ddk001]]
+Obviously, <math>y</math> is <math>3</math> times a power of <math>2</math>. Testing, we see <math>y=6</math> satisfy the equation so <math>a=2^{\frac{3}{64}}</math>. Therefore, <math>x=2^{192} \equiv \boxed{896} \pmod{1000}</math> ~[[Ddk001]]
-== Alternate solution ==
+== Alternate solution 1 ==
 If you've found <math>x</math> but you don't know that much number theory.
@@ Line 85: / Line 85: @@
 <cmath>2^{192} \equiv 336*336 \equiv \boxed{896}\bmod 1000</cmath>
+== Alternate solution 2 ==
+Another way to find <math>x \bmod 1000</math> using modular arithmetic.
+In the same way as solution <math>1</math>, we can find that.
+<cmath>x\equiv 21\bmod 125, x\equiv 0\bmod 8.</cmath>
+<cmath>x = 8m = 125n+21</cmath> For some positive integers <math>m</math> and <math>n</math>.
+Taking the equation mod <math>8</math> gives <cmath>5n+5 \equiv 0\bmod 8</cmath>
+<cmath>n \equiv 7\bmod 8</cmath>
+<cmath>n = 8k-1</cmath> For some positive integer <math>k</math>.
+Plug this back into the original equation.
+<cmath>8m = 125(8k-1)+21</cmath>
+<cmath>8m = 1000k-104</cmath>
+<cmath>x = 8m = 1000k - 104</cmath>
+<cmath>x \equiv -104 \equiv 896\bmod 1000</cmath>
+<cmath>x \equiv 896\bmod 1000</cmath>
+~sdfgfjh
 ==Video Solution by mop 2024==
 https://youtu.be/E-7YQ9ND5Ms

2017 AIME I (Problems • Answer Key • Resources)
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