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

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==Problem==
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lo
Find the number of elements in the set
 
{(a, b) ∈ N : 2 ≤ a, b ≤ 2023, loga
 
(b) + 6 logb
 
(a) = 5}
 
  
 
==Solution==
 
==Solution==

Revision as of 09:09, 27 October 2023

lo

Solution

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*} Solving the system gives $x = 4$ and $b = \frac{5}{4}$. Therefore, \[n = b^x = \frac{625}{256}.\] Therefore, the answer is $625 + 256 = \boxed{881}$.

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

Solution2

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

https://youtu.be/U96XHH23zhA

~IceMatrix

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

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