Difference between revisions of "2023 AIME I Problems/Problem 2"
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==Solution 3 (quick)== | ==Solution 3 (quick)== | ||
− | We can let <math>n=b^{4x^2}</math>. Then, in the first equation, the LHS becomes 2x and the RHS becomes 2x^2. Therefore, x must be 1 (x can't be 0). So now we know <math>n=b^4</math>. So we can plug this into the second equation to get. This gives <math>b\cdot4=5</math>, so <math>b= \frac{5}{4}</math> and <math>n= b^4=\frac{625}{256}</math>. Adding the numerator and denominator gives <math>\boxed{881}</math>. | + | We can let <math>n=b^{4x^2}</math>. Then, in the first equation, the LHS becomes <math>2x</math> and the RHS becomes <math>2x^2</math>. Therefore, <math>x</math> must be <math>1</math> (<math>x</math> can't be <math>0</math>). So now we know <math>n=b^4</math>. So we can plug this into the second equation to get. This gives <math>b\cdot4=5</math>, so <math>b= \frac{5}{4}</math> and <math>n= b^4=\frac{625}{256}</math>. Adding the numerator and denominator gives <math>\boxed{881}</math>. |
Honestly this problem is kinda misplaced. | Honestly this problem is kinda misplaced. | ||
~yrock | ~yrock | ||
− | |||
==Video Solution by TheBeautyofMath== | ==Video Solution by TheBeautyofMath== |
Revision as of 18:34, 23 May 2024
Contents
[hide]Problem
Positive real numbers and
satisfy the equations
The value of
is
where
and
are relatively prime positive integers. Find
Video Solution & More by MegaMath
https://www.youtube.com/watch?v=jxY7BBe-4gU
Solution 1
Denote .
Hence, the system of equations given in the problem can be rewritten as
Solving the system gives
and
.
Therefore,
Therefore, the answer is
.
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Solution 2
We can use the property that on the first equation. We get
. Then, subtracting
from both sides, we get
, therefore
. Substituting that into our first equation, we get
. Squaring, reciprocating, and simplifying both sides, we get the quadratic
. Solving for
, we get
and
. Since the problem said that
,
. To solve for
, we can use the property that
.
, so
. Adding these together, we get
~idk12345678
Solution 3 (quick)
We can let . Then, in the first equation, the LHS becomes
and the RHS becomes
. Therefore,
must be
(
can't be
). So now we know
. So we can plug this into the second equation to get. This gives
, so
and
. Adding the numerator and denominator gives
.
Honestly this problem is kinda misplaced.
~yrock
Video Solution by TheBeautyofMath
~IceMatrix
See also
2023 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
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.