3^(2p) + 3^(3q) + 3^(5r) = 3^(7s)
Find the minimum value of p+q+r+s
where p,q,r,s are all positive integer.

- anonymous

- katieb

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- mayankdevnani

9^p+27^q+3^5r=3^7s

- mayankdevnani

ok

- mayankdevnani

got it @sauravshakya

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## More answers

- anonymous

What was that?

- mayankdevnani

we just square and cube the value

- anonymous

The question is to find minimum value of p+q+r+s

- mayankdevnani

then common 3 as a base

- mayankdevnani

and then add powers

- mayankdevnani

then you solve it

- anonymous

I have never seen that rule.

- mayankdevnani

ok so go through wolframalpha

- mayankdevnani

http://www.wolframalpha.com/input/?i=3%5E%282p%29+%2B+3%5E%283q%29+%2B+3%5E%285r%29+%3D+3%5E%287s%29+Find+the+minimum+value+of+p%2Bq%2Br%2Bs

- anonymous

I dont think that will also help here.

- mayankdevnani

go through

- anonymous

There is nothing.

- anonymous

I am seeing it.

- anonymous

So, what is the answer?

- anonymous

Give a min, but at tell u the answer can't be find the above way.

- anonymous

What I figured out till now is s must be an odd number....

- anonymous

@experimentX PLZ see this

- experimentX

this doesn't look nice ...
saying that find the minimum value of p+q+r+s <--- this should be close to first instances.

- anonymous

Cant we find that......

- experimentX

exponential diophantine equations are awful!! you should probably tag mukushla

- anonymous

This question is not of mine........ Someone posted this today and said a 8th grade challenging question.

- anonymous

But he is not online.

- anonymous

@mukushla

- experimentX

you don't do diophantine equations on 8th grade.

- anonymous

Even I dont know that.

- anonymous

This surely cant be a 8th grade question.

- mayankdevnani

@mathslover i think he do this

- mayankdevnani

very dhinchach question

- anonymous

Now, what I am trying to PROVE is 3^(2p) + 3^(3q) + 3^(5r) = 3^(7s) is never true for any value of p,q,r,s

- experimentX

anyway this was the output I got from mathematica
\[ \left\{\left\{p\to \frac{8}{5}+2 i,q\to \frac{223}{10}+\frac{53 i}{5},r\to -\frac{71}{10}-\frac{49 i}{10},\\ s\to \frac{180 i \pi +\text{Log}\left[3^{-\frac{71}{2}-\frac{49 i}{2}}+3^{\frac{16}{5}+4 i}+3^{\frac{669}{10}+\frac{159 i}{5}}\right]}{7 \text{Log}[3]}\right\}\right\} \]
I trolled enough ... not time to go strolling!!

- anonymous

I dont know it even has a solution.

- experimentX

*now ..
seeya later!!

- anonymous

ok...... BYE thanx for trying.

- experimentX

call it hardly a try .. hehehe

- shubhamsrg

how about this :
i dont know if this'll work :
( 3^(2p) + 3^(3q) + 3^(5r) )/3 >= 3^( (2p + 3q + 5r)/3 )
or
3^(7s-1) >= 3^( (2p + 3q + 5r)/3 ) ??

- shubhamsrg

@ujjwal p,q,r,s are only positive integers..

- anonymous

@shubhamsrg
( 3^(2p) + 3^(3q) + 3^(5r) )/3 >= 3^( (2p + 3q + 5r)/3 )
HOW???

- shubhamsrg

from AM>=GM

- anonymous

Oh >= got it.

- anonymous

Yeah nice one.

- shubhamsrg

and continuing that,
we have
21s >= 2p + 3q + 5r + 3
but still no idea if that'll help..

- anonymous

And I got:
|dw:1349955495301:dw|

- anonymous

I guess NOW, we can prove it has no SOLUTION

- shubhamsrg

hmm ??

- hartnn

*

- anonymous

THAT GIVES:
|dw:1349955687180:dw|

- anonymous

Now, |dw:1349955794330:dw|

- anonymous

Now, can I say No solution...... Or I lack something.

- shubhamsrg

well you have made an assumption that 5r<3q ..
only that makes LHS an integer..otherwise you cant say..

- shubhamsrg

am sorry,,you assumed 5r>3q ..

- shubhamsrg

but conclusion to my saying is that you made an assumtion..

- anonymous

oh...... no assumtion.

- anonymous

L.H.S or R.H.S can be both negative

- anonymous

Just looking for integer.......
L.H.S will always be integer
R.H.S will always be in decimal.

- shubhamsrg

LHS is not necessary an integer.. we only know that p,q,r,s are integers..

- shubhamsrg

and agreed that RHS will always be non-integer..

- anonymous

Oh ya....... got it @shubhamsrg

- anonymous

So,from there we conclude that 5r<3q

- shubhamsrg

guess not,,how can you say that ?

- anonymous

Because if 5r>=3q then R.H.S will always be integer.

- anonymous

I mean L.H.S

- shubhamsrg

but there is also 5r<=3q possibility ,,
the ques doesnt say that 5r - 3q > 0 /// even if that's true..the proof you mention is invalid..

- anonymous

oh wait........ I guess R.H.S can be integer too.

- shubhamsrg

also, we may always conclude that 7s > 2p.3q,5r ..but nothing can be said for inequalities in between 2p,3q and 5r..

- anonymous

Ok further simplification of it gives:
|dw:1349956922192:dw|

- anonymous

divide both sides by \(3^{2p}\)\[1+3^{3q-2p}+3^{5r-2p}=3^{7s-2p}\]right hand side is a power of 3 so does the left hand side so we must have\[2p=3q=5r\]\[7s-2p=1\]

- anonymous

GREAT.

- anonymous

Now, we need to find its minimum value.

- anonymous

what is that fraction?

- anonymous

p+q+r+s = (207p+15)/105

- anonymous

right?

- anonymous

ahh yeah :)

- anonymous

Now, we need to find its minimum integer value.

- anonymous

Oh wait...... But that can make other numbers decimal.

- anonymous

u better start with \[7s-2p=1\]ans\[p=\frac{7s-1}{2}\]its immidiate that s is odd let s=2m+1\[p=7m+3\]now find the smallest value of m for which p is a multiple of 3 and 5

- anonymous

if u wanna check the answer i found is 106 for
\[m=6\]\[p=45\]\[q=30\]\[r=18\]\[s=13\]

- anonymous

Thanx @mukushla u r really amazing.

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