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>. | ||
− | ==Solution== | + | ==Solution 1== |
The first condition implies | The first condition implies | ||
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Using [[Chinese Remainder Theorem]], we get that <math>x\equiv \boxed{896}\bmod 1000</math>, finishing the solution. | Using [[Chinese Remainder Theorem]], we get that <math>x\equiv \boxed{896}\bmod 1000</math>, finishing the solution. | ||
+ | |||
+ | ==Solution 2 (Another way to find <math>a</math>)== | ||
+ | |||
+ | <math>\log_a(\log_a(\log_a 2) + \log_a 24 - 128) = 128</math> | ||
+ | |||
+ | <math>\implies \log_a(\log_a 2))+log_a(24)=a^{128}+128</math> | ||
+ | |||
+ | <math>\implies \log_a(\log_a 2^{24})=a^{128}+128</math> | ||
+ | |||
+ | <math>\implies 2^{24}=a^{a^{(a^{128}+128)}}</math> | ||
+ | |||
+ | 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> | ||
+ | |||
+ | <math>\implies 24=\frac{y}{128} \cdot 2^{y \cdot (2^{y-7}+1)}</math> | ||
+ | |||
+ | <math>\implies 3 \cdot 2^{10}=y \cdot 2^{y \cdot (2^{y-7}+1)}</math> | ||
+ | |||
+ | 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> | ||
== Alternate solution == | == Alternate solution == |
Revision as of 14:29, 3 January 2024
Contents
Problem 14
Let and
satisfy
and
. Find the remainder when
is divided by
.
Solution 1
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 Chinese Remainder Theorem, we get that , finishing the solution.
Solution 2 (Another way to find
)
Obviously letting will simplify a lot and to make the
term simpler, let
. Then,
Obviously, y is times a power of
. (It just makes sense.) Testing, we see
satisfy the equation so
. Therefore,
Alternate solution
If you've found but you don't know that much number theory.
Note , so what we can do is take
and keep squaring it (mod 1000).
Video Solution by mop 2024
~r00tsOfUnity
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.