Math olympiad products question

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Homework Help Overview

The problem involves finding the smallest integer n such that the product \(5 (32 + 22)(34 + 24)(38 + 28)...(32n + 22n) > 9256\). This falls under the subject area of inequalities and products in mathematics, particularly in the context of competitive mathematics such as math olympiads.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss various approaches to simplify the product and explore the implications of ignoring certain terms. There are attempts to apply the AM-GM inequality and logarithmic properties to establish bounds. Some participants question the validity of their assumptions and the significance of correction factors in their calculations.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem and sharing insights. Some guidance has been offered regarding the application of inequalities and the need for closed forms, but there is no explicit consensus on the final approach or solution.

Contextual Notes

Participants note constraints such as the need to find a lower bound for the product and the implications of discarding certain terms. There is also mention of conflicting information regarding the correct value of n, with some sources suggesting different answers.

  • #31
haruspex said:
That n is not the same as the n in the problem.
oh you careless me but i am still getting a different result this is my working

##
3^{2^{n+1}} ( 1- (\frac{2}{3})^{2^{n+1} -1 }) \\
3^{2^{n+1}} - 3( 2^{2^{n+1} -1 })

##
now what am i doing wrong
 
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  • #32
vishnu 73 said:
now what am i doing wrong
You are not applying the upper bound correctly in the sum. If the sum is up to the term rk then the numerator should have 1-rk+1. You omitted the +1.
 
  • #33
oh wait the index starts at i=0 not 1 so it must be 22n+1 correct ?

then it is equal to
##
3^{2^{n+1}} - 2^{2^{n+1}}
##
 
  • #34
vishnu 73 said:
oh wait the index starts at i=0 not 1 so it must be 22n+1 correct ?

then it is equal to
##
3^{2^{n+1}} - 2^{2^{n+1}}
##
Yes.
 
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