Difference between revisions of "2023 AIME I Problems/Problem 2"

(Created page with "__TOC__ ==Problem== Problem statement ==Solutions== ===Solution 1=== Solution by someone ===Solution 2=== Solution by someone else ==See also== {{AIME box|year=2023|num-b=1...")
 
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==Solutions==
 
==Solutions==
 
===Solution 1===
 
===Solution 1===
Solution by someone
+
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 . \
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bx & = 1 + x .
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\end{align*}
 +
</cmath>
 +
 
 +
Thus, <math>x = 4</math> and <math>b = \frac{5}{4}</math>.
 +
Therefore,
 +
<cmath>
 +
\begin{align*}
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n & = b^x \
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& = \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 $x = \log_b n$. Hence, the system of equations given in the problem can be rewritten as \begin{align*} \sqrt{x} & = \frac{1}{2} x . \\ bx & = 1 + x . \end{align*}

Thus, $x = 4$ and $b = \frac{5}{4}$. Therefore, \begin{align*} n & = b^x \\ & = \frac{625}{256} . \end{align*}

Therefore, the answer is $625 + 256 = \boxed{\textbf{(881) }}$.

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Solution 2

Solution by someone else

See also

2023 AIME I (ProblemsAnswer KeyResources)
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