Difference between revisions of "2023 AIME I Problems/Problem 2"
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<cmath> | <cmath> | ||
\begin{align*} | \begin{align*} | ||
− | \sqrt{x} & = \frac{1}{2} x | + | \sqrt{x} & = \frac{1}{2} x , \ |
bx & = 1 + x . | bx & = 1 + x . | ||
\end{align*} | \end{align*} | ||
</cmath> | </cmath> | ||
− | |||
Thus, <math>x = 4</math> and <math>b = \frac{5}{4}</math>. | Thus, <math>x = 4</math> and <math>b = \frac{5}{4}</math>. | ||
Therefore, | Therefore, | ||
− | <cmath> | + | <cmath>n = b^x = \frac{625}{256}.</cmath> |
− | |||
− | n | ||
− | |||
− | |||
− | </cmath> | ||
− | |||
Therefore, the answer is <math>625 + 256 = \boxed{881}</math>. | Therefore, the answer is <math>625 + 256 = \boxed{881}</math>. | ||
Revision as of 13:35, 8 February 2023
Contents
[hide]Problem
Positive real numbers and satisfy the equations The value of is where and are relatively prime positive integers. Find
Solution 1
Denote . Hence, the system of equations given in the problem can be rewritten as Thus, and . Therefore, Therefore, the answer is .
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Solution 2
Solution by someone else
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 |