What is the smallest n for a small inequality?

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Discussion Overview

The discussion revolves around finding the smallest integer \( n \) that satisfies a specific inequality involving logarithms and binomial coefficients. The conversation includes both theoretical considerations and numerical methods for solving the problem.

Discussion Character

  • Exploratory
  • Mathematical reasoning

Main Points Raised

  • One participant asks for the smallest \( n \) such that \( \lg {n \choose 0.15n} + 0.15n \geq 112 \).
  • Another participant presents a slightly different inequality, asking for the smallest \( n \) such that \( \lg {n \choose 0.15n} + {n \choose 0.15n} \geq 112 \), and inquires about the meaning of \( \lg \).
  • Several participants clarify that \( \lg \) refers to logarithm base 2.
  • There is a question regarding whether the logarithm includes the sum of both combinations, leading to a correction of the initial formulation of the inequality.
  • A participant suggests that it might be easier to approach the problem numerically rather than analytically.
  • Another participant proposes an equation \( 2^{LH} = 2^{RH} \) but agrees that a numerical method may be the best approach.

Areas of Agreement / Disagreement

Participants express uncertainty about the formulation of the inequality and whether the logarithm encompasses both combinations. There is no consensus on the best method to solve the problem, with some favoring numerical methods while others consider analytical approaches.

Contextual Notes

The discussion includes potential limitations regarding the clarity of the inequality formulation and the assumptions about the logarithmic expression. The exact mathematical steps for solving the inequality remain unresolved.

Dragonfall
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What is the smallest n such that

[tex]\lg {n\choose0.15n} + 0.15n \geq {112}[/tex]
 
Last edited:
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Dragonfall said:
What is the smallest n such that

[tex]\lg {n\choose0.15n} + {n\choose0.15n} \geq {112}[/tex]
What is lg?
 
Math_QED said:
What is lg?

Log base 2
 
Dragonfall said:
Log base 2
Does the log contain the sum of both combinations? (Then you should have added more brackets)
 
Math_QED said:
Does the log contain the sum of both combinations? (Then you should have added more brackets)

Sorry I formulated the inequality wrong. It's fixed now.

The log only contains the binomial.

I think it might be easier just to do this numerically...
 
You can try to do:

2^LH = 2^RH

But I think the best approach is a numerical method.
 

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