Difference between revisions of "2023 AIME I Problems/Problem 2"
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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
− | == | + | ==Solution 2== |
− | Denote b=n^x. | + | Denote b=n^x. Hence, the system of equations given in the problem can be rewritten as |
sqrt(x)=x/2, b*x=1+x | sqrt(x)=x/2, b*x=1+x | ||
Thus, x=x^2/4, x=4. So, n=b^4 | Thus, x=x^2/4, x=4. So, n=b^4 |
Revision as of 23:59, 11 December 2023
Problem
Positive real numbers and
satisfy the equations
The value of
is
where
and
are relatively prime positive integers. Find
Solution
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
Denote b=n^x. Hence, the system of equations given in the problem can be rewritten as sqrt(x)=x/2, b*x=1+x Thus, x=x^2/4, x=4. So, n=b^4 Then, 4b=1+4. So, b=5/4. Then, n=625/256 Ans=881
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.