Difference between revisions of "2017 AIME I Problems/Problem 14"
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Let <math>a > 1</math> and <math>x > 1</math> satisfy <math>\log_a(\log_a(\log_a 2) + \log_a 24 - 128) = 128</math> and <math>\log_a(\log_a x) = 256</math>. Find the remainder when <math>x</math> is divided by <math>1000</math>. | Let <math>a > 1</math> and <math>x > 1</math> satisfy <math>\log_a(\log_a(\log_a 2) + \log_a 24 - 128) = 128</math> and <math>\log_a(\log_a x) = 256</math>. Find the remainder when <math>x</math> is divided by <math>1000</math>. | ||
Revision as of 19:33, 8 March 2017
Problem 14
Let 
and 
satisfy 
and 
. Find the remainder when 
is divided by 
.
Solution
The first condition implies
![\[a^{128} = \log_a\log_a 2 + \log_a 24 - 128\]](http://latex.artofproblemsolving.com/3/7/e/37e141af144886175e8f888317b997783ed8351c.png)
![\[128+a^{128} = \log_a\log_a 2^{24}\]](http://latex.artofproblemsolving.com/a/a/f/aaf09086fc61ef1ac7826a4e494d7191da4210b8.png)
![\[a^{a^{128}a^{a^{128}}} = 2^{24}\]](http://latex.artofproblemsolving.com/9/9/d/99d0629298cfc70e32f871a4c407d469282ea98a.png)
![\[\left(a^{a^{128}}\right)^{\left(a^{a^{128}}\right)} = 2^{24} = 8^8\]](http://latex.artofproblemsolving.com/3/f/2/3f23eccc1a1f771c12339626fa20390f60b1a790.png)
So 
.
Putting each side to the power of 
:
![\[\left(a^{128}\right)^{\left(a^{128}\right)} = 8^{128} = 64^{64},\]](http://latex.artofproblemsolving.com/9/7/f/97f33b05e81a42b56c7513b809b3c9a4449181f3.png)
so 
. Specifically,
![\[\log_a(y) = \frac{\log_2(y)}{\log_2(a)} = \frac{64}{3}\log_2(y)\]](http://latex.artofproblemsolving.com/1/a/4/1a44c3a772024f44784477de5bdfb2d64fe0a2bf.png)
so we have that
![\[256 = \log_a(\log_a(y)) = \frac{64}{3}\log_2\left(\frac{64}{3}\log_2(x)\right)\]](http://latex.artofproblemsolving.com/5/e/6/5e6e335f6eca89a2263d9edb91f2dce98d6079ea.png)
![\[12 = \log_2\left(\frac{64}{3}\log_2(x)\right)\]](http://latex.artofproblemsolving.com/c/5/5/c55cc4782cbd511fe1cb8702814856a87505e0a2.png)
![\[2^{12} = \frac{64}{3}\log_2(x)\]](http://latex.artofproblemsolving.com/c/2/e/c2e031ac284a023d2bdafab29e41a312b4ad0093.png)
![\[192 = \log_2(x)\]](http://latex.artofproblemsolving.com/1/1/d/11da6c5f64d54c0fcf9012af6060fd6d91310927.png)
![\[x = 2^{192}\]](http://latex.artofproblemsolving.com/2/5/6/256863c8c0cd4c0b9fc75e2ecc1c493d69c2b334.png)
We only wish to find 
. To do this, we note that 
and now, by the Chinese Remainder Theorem, wish only to find 
. By Euler's Theorem:
![\[2^{\phi(125)} = 2^{100} \equiv 1\bmod 125\]](http://latex.artofproblemsolving.com/8/8/7/8873102edbcf210b4457a0e3ba46e7daaa4ce2d6.png)
so
![\[2^{192} \equiv \frac{1}{2^8} \equiv \frac{1}{256} \equiv \frac{1}{6} \bmod 125\]](http://latex.artofproblemsolving.com/5/e/d/5ed2fd9516b24a348efdddd3db1d359e8ec46cd2.png)
so we only need to find the inverse of 
. It is easy to realize that 
, so
![\[x\equiv 21\bmod 125, x\equiv 0\bmod 8.\]](http://latex.artofproblemsolving.com/4/4/d/44dfd8d02e0ce83290b0d7ddaad6b4280ad5e4d6.png)
Using CRT, we get that 
, finishing the solution.
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
