2023 AIME I Problems/Problem 2: Difference between revisions
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==Solutions== | ==Solutions== | ||
===Solution 1=== | ===Solution 1=== | ||
Denote <math>x = \log_b n</math>. | |||
Hence, the system of equations given in the problem can be rewritten as | |||
<cmath> | |||
\begin{align*} | |||
\sqrt{x} & = \frac{1}{2} x . \\ | |||
bx & = 1 + x . | |||
\end{align*} | |||
</cmath> | |||
Thus, <math>x = 4</math> and <math>b = \frac{5}{4}</math>. | |||
Therefore, | |||
<cmath> | |||
\begin{align*} | |||
n & = b^x \\ | |||
& = \frac{625}{256} . | |||
\end{align*} | |||
</cmath> | |||
Therefore, the answer is <math>625 + 256 = \boxed{\textbf{(881) }}</math>. | |||
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | |||
===Solution 2=== | ===Solution 2=== | ||
Revision as of 12:59, 8 February 2023
Problem
Problem statement
Solutions
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 | ||
