2023 AIME I Problems/Problem 2

Revision as of 13:24, 8 February 2023 by Mrthinker (talk | contribs) (Solution 1)

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 $625 + 256 = \boxed{881}$.

~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
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All AIME Problems and Solutions