Difference between revisions of "2017 AIME I Problems/Problem 14"
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<cmath>12 = \log_2\left(\frac{64}{3}\log_2(x)\right)</cmath> | <cmath>12 = \log_2\left(\frac{64}{3}\log_2(x)\right)</cmath> | ||
− | <cmath>2^12 = \frac{64}{3}\log_2(x)</cmath> | + | <cmath>2^{12} = \frac{64}{3}\log_2(x)</cmath> |
<cmath>192 = \log_2(x)</cmath> | <cmath>192 = \log_2(x)</cmath> | ||
Line 49: | Line 49: | ||
<cmath>x\equiv 21\bmod 125, x\equiv 0\bmod 8.</cmath> | <cmath>x\equiv 21\bmod 125, x\equiv 0\bmod 8.</cmath> | ||
− | Using CRT, we get that <math>x\equiv \boxed{896}\bmod 1000</math>, finishing the solution. | + | Using CRT, we get that <math>x\equiv \boxed{896}\bmod 1000</math>, finishing the solution. |
== See also == | == See also == | ||
{{AIME box|year=2017|n=I|num-b=13|num-a=15}} | {{AIME box|year=2017|n=I|num-b=13|num-a=15}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 16:32, 8 March 2017
Problem
Let and satisfy and . Find the remainder when is divided by .
Solution
The first condition implies
So .
Putting each side to the power of :
so . Specifically,
so we have that
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:
so
so we only need to find the inverse of . It is easy to realize that , so
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.