Melody

Yes I get that, and it is good logic,  but how did you know that k was defined at all. 

OR maybe k is 12 and then there would not be a second k value.

....

I guess it was a reasonable assumtion that k was defined....and that it was not 12.

And yes if any such k exists (which I know it does becasue I worked it out ) and k is not 12, then there will be 2  k values, and their sum is 24 just as you have said.    

Builderboi, Your answer is correct but i had to think really hard before I could see the relevance of your explanation.

That is really good to here.  Thanks  

And thanks for your teaching style answer Builderboi.  

If you become a member you will get known (positively I think) and then you will get a consistently good response (like this one)  from answerers here.  

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.

How about you write something of what you have done for yourself.

It seems like you just want to copy an answer.

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

break the mods into their prime factors

6=2*3

7

8=2*2*2

9=3*3

So  6,7,8 and 9 will all go into  2*3*7*2*2*3=  6*7*4*3 = 504

There will be 1 solution less than 504 and then 1 solution for each following product of 504

6*7*8*9 = [6*7*4*3]* 6

So there will be 6 solutions less than   6*7*8*9

What is your guess ?

Have you tried some different possibilities?

Whawt have you tried?

A reasonable guess Pureant but unfortunately it is not correct.   

* I will give you thumbs up for your attempt ;)

Think about normal numbers

The multiplication inverse of  4 is 1/4  .

This is becasue  4 * 1/4 = 1

It is the same with modular arithemetic.

The number that is the inverse of 4 is the one that can be multipled by 4 to get 1.

So the inverse of 4 mod 35   is  B such that    4B=1  mod35

4*9=36 which is 1 mod 35

therefore the inverse of 4 is 9 and the inverse of 9 is 4  mod35

so the question becomes  9*4 = 36  = 1 mod35

It is about as correct as your question.

Garbage in, garbage out