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Let \(P = 5^{1/5} \cdot 25^{1/25} \cdot 125^{1/125} \cdot 625^{1/625} \dotsm\)
Then \(P\) can be expressed in the form \(a^{b/c}\) where \(a,b,\) and \(c\) are positive integers. Find the smallest possible value of \(a+b+c\).

 

So far I know that a general expression for each factor is 5^(n/5^n) and a=5. can i get guidance to continue?

 Dec 10, 2021
edited by Guest  Dec 10, 2021
 #1
avatar+132 
+1

Very good question.

If we express everything in terms of roots, we get \(P = \sqrt[5]{5} \cdot \sqrt[25]{25} \cdot \sqrt[125]{125} \cdot \dots \)

\(P = \sqrt[5]{5} \cdot \sqrt[5]{\sqrt[5]{25}} \cdot \sqrt[5]{\sqrt[5]{\sqrt[5]{125}}} \cdot \dots\)

Let 5th root of 5 = x.

\(P = x \cdot \sqrt[5]{x^2} \cdot \sqrt[5]{\sqrt[5]{x^3}} \cdot \dots\)

Try to base your solution off of the fact that \(P^{1/5} = x^{1/5} \cdot x^{2/25} \cdot x^{3/125} \cdot \dots\).

If you need more help, just reply.

You are correct that a=5.

 Dec 10, 2021
edited by tinfoilhat  Dec 10, 2021
edited by tinfoilhat  Dec 10, 2021
 #4
avatar+115901 
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Attention:  Tinfoilhat

 

I know that you have given some good answers and I thank you for those...  but I think in this case you did not know how to answer this question. 

 

I encourage people to leave part answers but it is reasonable for the asker to assume that you do actually know how to finish it.

 

If you have not finished the question for yourself (maybe because of time restraints), or you do not know how to finish it, please make this very clear to the asker.

Melody  Dec 18, 2021
edited by Melody  Dec 18, 2021
 #2
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+1

I might need more help actually.

 

I found that a=5, but where should I go from here.. 

 Dec 18, 2021
edited by Guest  Dec 18, 2021
 #3
avatar+115901 
+2

 I like this question too.

 

\(25^\frac{1}{25}=5^{\frac{2}{5^2}}\\~\\ 125^\frac{1}{125}=5^{\frac{3}{5^3}}\\~\\ 625^\frac{1}{625}=5^{\frac{4}{5^4}}\\~\\ \)

\(P = 5^{1/5} \cdot 25^{1/25} \cdot 125^{1/125} \cdot 625^{1/625} \dotsm\\~\\ P = 5^{1/5} \cdot 5^{2/5^2} \cdot 5^{3/5^3} \cdot 5^{4/5^4} \dotsm\\~\\ P = 5^{{1/5}+2/5^2+{3/5^3} +{4/5^4} }\dotsm\\~\\ \)

So I need to find a sum to infinity.

 

\(\frac{1}{5}+\frac{2}{5^2}+\frac{3}{5^3}+\dots +\\~\\ =\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+\dots +\frac{1}{5^2}+\frac{1}{5^3}+\frac{1}{5^4}\dots +\frac{1}{5^3}+\frac{1}{5^4}+\frac{1}{5^5}\dots +\\~\\ =(\text{infinite sum of GP a=1/5, r=1/5})\dots \\ \qquad+(\text{infinite sum of GP a=1/25, r=1/5})\dots \\ \qquad+(\text{infinite sum of GP a=1/125, r=1/5})\dots +\\~\\ =\left[\frac{1}{5}\div(1-\frac{1}{5}))\right]+\left[\frac{1}{25}\div(1-\frac{1}{5}))\right]+\left[\frac{1}{125}\div(1-\frac{1}{5}))\right]+\dots \\ =\left[\frac{1}{5}\times\frac{5}{4}\right]+\left[\frac{1}{25}\times\frac{5}{4}\right]+\left[\frac{1}{125}\times\frac{5}{4}\right]+\dots \\ =\frac{1}{4}+\frac{1}{20}+\frac{1}{100}\dots\\~\\ \text{Infinite sum of a GP a=1/4 r=1/5}\\ =\frac{1}{4}\div(1-\frac{1}{5})\\~\\ =\frac{1}{4}\times \frac{5}{4}\\~\\ =\frac{5}{16} \)

 

 

\(P = 5^{1/5} \cdot 25^{1/25} \cdot 125^{1/125} \cdot 625^{1/625} \dotsm\\ P= 5^\frac{5}{16}\)

 

NOTE: 

I wrote the latex and worked the logic at the same time so there is plenty of room for careless error.

You need to work through (understand) what I have done and find any errors that might be there.

 

Unless you only want the answer of course in which case you may or may not get lucky.

 

 

LaTex:

\frac{1}{5}+\frac{2}{5^2}+\frac{3}{5^3}+\dots +\\~\\
=\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+\dots +\frac{1}{5^2}+\frac{1}{5^3}+\frac{1}{5^4}\dots +\frac{1}{5^3}+\frac{1}{5^4}+\frac{1}{5^5}\dots +\\~\\
=(\text{infinite sum of GP a=1/5, r=1/5})\dots \\
\qquad+(\text{infinite sum of GP a=1/25, r=1/5})\dots \\
\qquad+(\text{infinite sum of GP a=1/125, r=1/5})\dots +\\~\\
=\left[\frac{1}{5}\div(1-\frac{1}{5}))\right]+\left[\frac{1}{25}\div(1-\frac{1}{5}))\right]+\left[\frac{1}{125}\div(1-\frac{1}{5}))\right]+\dots \\

=\left[\frac{1}{5}\times\frac{5}{4}\right]+\left[\frac{1}{25}\times\frac{5}{4}\right]+\left[\frac{1}{125}\times\frac{5}{4}\right]+\dots \\

=\frac{1}{4}+\frac{1}{20}+\frac{1}{100}\dots\\~\\
\text{Infinite sum of a GP a=1/4   r=1/5}\\
=\frac{1}{4}\div(1-\frac{1}{5})\\~\\
=\frac{1}{4}\times \frac{5}{4}\\~\\
=\frac{5}{16}
 

 

=

 Dec 18, 2021

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