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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.

 May 16, 2022
 #1
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+1

\(P = 5^{4/13}\)

 

Therefore, a + b + c = 5 + 4 + 13 = 22.

 May 16, 2022
 #2
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+1

Simplify the following:
5^(1/5) 25^(1/25) 125^(1/125) 625^(1/625)

625^(1/625) = (5^4)^(1/625):


5^(1/5) 25^(1/25) 125^(1/125)×5^(4/625)

125^(1/125) = (5^3)^(1/125):


5^(1/5) 25^(1/25)×5^(3/125)×5^(4/625)

25^(1/25) = (5^2)^(1/25):


5^(1/5)×5^(2/25)×5^(3/125)×5^(4/625)

5^(1/5)×5^(2/25)×5^(3/125)×5^(4/625) = 5^(1/5 + 2/25 + 3/125 + 4/625):


5^(1/5 + 2/25 + 3/125 + 4/625)

Put 1/5 + 2/25 + 3/125 + 4/625 over the common denominator 625. 1/5 + 2/25 + 3/125 + 4/625 = 125/625 + (25×2)/625 + (5×3)/625 + 4/625:


5^(125/625 + (25×2)/625 + (5×3)/625 + 4/625)

25×2 = 50:
5^(125/625 + 50/625 + (5×3)/625 + 4/625)

5×3 = 15:


5^(125/625 + 50/625 + 15/625 + 4/625)

125/625 + 50/625 + 15/625 + 4/625 = (125 + 50 + 15 + 4)/625:


5^((125 + 50 + 15 + 4)/625)

125 + 50 + 15 + 4 = 194:

5^(194/625)

 May 16, 2022
 #3
avatar+118146 
+2

Thanks guests :)

 

P = 5^{1/5} \cdot 25^{1/25} \cdot 125^{1/125} \cdot 625^{1/625} \dotsm

 

\(P = 5^{1/5} \cdot 25^{1/25} \cdot 125^{1/125} \cdot 625^{1/625} \dotsm\\ P = 5^{1/5} \cdot 5^{2/5^2} \cdot 5^{3/5^3} \cdot 5^{4/5^4} \dotsm\\ P = 5^{(\frac{1}{5}+\frac{2}{5^2}+\frac{3}{5^3}+\frac{4}{5^4}\dotsm\\)} \\ log_5 P = {(\frac{1}{5}+\frac{2}{5^2}+\frac{3}{5^3}+\frac{4}{5^4}\dotsm)} \\ log_5 P = {(\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}\dotsm)} + {(\frac{1}{5^2}+\frac{1}{5^3}+\frac{1}{5^4}\dotsm)} +{(\frac{1}{5^3}+\frac{1}{5^4}+\frac{1}{5^5}\dotsm)+ \dots}\\ log_5 P =(\frac{1}{5}\div \frac{4}{5})+(\frac{1}{25}\div \frac{4}{5})+(\frac{1}{125}\div \frac{4}{5})+ \dots\\ log_5 P =(\frac{1}{5}\times \frac{5}{4})+(\frac{1}{25}\times\frac{5}{4})+(\frac{1}{125}\times\frac{5}{4})+ \dots\\ log_5 P =(\frac{1}{4})+(\frac{1}{20})+(\frac{1}{100})+ \dots\\ log_5 P =\frac{1}{4}\div \frac{4}{5}\\ log_5 P =\frac{1}{4}\div \frac{4}{5}\\ log_5 P =\frac{1}{4}\times \frac{5}{4}\\ log_5 P =\frac{5}{16}\\ P=5^{\frac{5}{16}}\\ P\approx 1.65359 \)

 

 

LaTex

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^{(\frac{1}{5}+\frac{2}{5^2}+\frac{3}{5^3}+\frac{4}{5^4}\dotsm\\)} \\
log_5 P = {(\frac{1}{5}+\frac{2}{5^2}+\frac{3}{5^3}+\frac{4}{5^4}\dotsm)} \\

log_5 P = {(\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}\dotsm)} 
+ {(\frac{1}{5^2}+\frac{1}{5^3}+\frac{1}{5^4}\dotsm)}
+{(\frac{1}{5^3}+\frac{1}{5^4}+\frac{1}{5^5}\dotsm)+ \dots}\\

log_5 P =(\frac{1}{5}\div \frac{4}{5})+(\frac{1}{25}\div \frac{4}{5})+(\frac{1}{125}\div \frac{4}{5})+ \dots\\
log_5 P =(\frac{1}{5}\times \frac{5}{4})+(\frac{1}{25}\times\frac{5}{4})+(\frac{1}{125}\times\frac{5}{4})+ \dots\\
log_5 P =(\frac{1}{4})+(\frac{1}{20})+(\frac{1}{100})+ \dots\\
log_5 P =\frac{1}{4}\div \frac{4}{5}\\
log_5 P =\frac{1}{4}\div \frac{4}{5}\\
log_5 P =\frac{1}{4}\times \frac{5}{4}\\
log_5 P =\frac{5}{16}\\
P=5^{\frac{5}{16}}\\
P\approx 1.65359

 May 17, 2022
 #4
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+1

Hi Melody:

 

When multiplied together, you get the following number:

 

(5^(1/5)) * (25^(1/25)) * (125^(1/125)) * (625^(1/625)) ==1.64801169512666976301802035295......etc.

 

1 - Guest #1 answer ==5^(4/13) ==1.64084551246609057805181222444......etc.

 

2 - Guest #2 answer==5^(194/625)==1.64801169512666976301802035295....etc.

 

3 - Melody's answer ==5^(5/16) ==1.65359110076253520866303943290......etc.

 May 17, 2022
 #5
avatar+118146 
+1

My answer is    \(5^{(\frac{5}{16})}\)

Unless I made a careless error (which there is no evidence of) this is the exact answer.

 

The 2 guest answers are not correct because  they do not account for the ... after the 4th term.

There is an infinite number of terms here, not just 4.

Melody  May 17, 2022
edited by Melody  May 17, 2022

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