sauravshakya
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.
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mayankdevnani
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9^p+27^q+3^5r=3^7s
mayankdevnani
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ok
mayankdevnani
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got it @sauravshakya
sauravshakya
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What was that?
mayankdevnani
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we just square and cube the value
sauravshakya
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The question is to find minimum value of p+q+r+s
mayankdevnani
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then common 3 as a base
mayankdevnani
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and then add powers
mayankdevnani
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then you solve it
sauravshakya
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I have never seen that rule.
mayankdevnani
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ok so go through wolframalpha
sauravshakya
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I dont think that will also help here.
mayankdevnani
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go through
sauravshakya
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There is nothing.
Zekarias
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I am seeing it.
sauravshakya
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So, what is the answer?
Zekarias
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Give a min, but at tell u the answer can't be find the above way.
sauravshakya
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What I figured out till now is s must be an odd number....
sauravshakya
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@experimentX PLZ see this
experimentX
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this doesn't look nice ...
saying that find the minimum value of p+q+r+s <--- this should be close to first instances.
sauravshakya
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Cant we find that......
experimentX
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exponential diophantine equations are awful!! you should probably tag mukushla
sauravshakya
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This question is not of mine........ Someone posted this today and said a 8th grade challenging question.
sauravshakya
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But he is not online.
sauravshakya
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@mukushla
experimentX
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you don't do diophantine equations on 8th grade.
sauravshakya
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Even I dont know that.
sauravshakya
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This surely cant be a 8th grade question.
mayankdevnani
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@mathslover i think he do this
mayankdevnani
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very dhinchach question
sauravshakya
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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
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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!!
sauravshakya
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I dont know it even has a solution.
experimentX
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*now ..
seeya later!!
sauravshakya
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ok...... BYE thanx for trying.
experimentX
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call it hardly a try .. hehehe
shubhamsrg
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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
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@ujjwal p,q,r,s are only positive integers..
sauravshakya
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@shubhamsrg
( 3^(2p) + 3^(3q) + 3^(5r) )/3 >= 3^( (2p + 3q + 5r)/3 )
HOW???
shubhamsrg
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from AM>=GM
sauravshakya
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Oh >= got it.
sauravshakya
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Yeah nice one.
shubhamsrg
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and continuing that,
we have
21s >= 2p + 3q + 5r + 3
but still no idea if that'll help..
sauravshakya
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And I got:
|dw:1349955495301:dw|
sauravshakya
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I guess NOW, we can prove it has no SOLUTION
shubhamsrg
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hmm ??
hartnn
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*
sauravshakya
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THAT GIVES:
|dw:1349955687180:dw|
sauravshakya
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Now, |dw:1349955794330:dw|
sauravshakya
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Now, can I say No solution...... Or I lack something.
shubhamsrg
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well you have made an assumption that 5r<3q ..
only that makes LHS an integer..otherwise you cant say..
shubhamsrg
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am sorry,,you assumed 5r>3q ..
shubhamsrg
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but conclusion to my saying is that you made an assumtion..
sauravshakya
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oh...... no assumtion.
sauravshakya
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L.H.S or R.H.S can be both negative
sauravshakya
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Just looking for integer.......
L.H.S will always be integer
R.H.S will always be in decimal.
shubhamsrg
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LHS is not necessary an integer.. we only know that p,q,r,s are integers..
shubhamsrg
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and agreed that RHS will always be non-integer..
sauravshakya
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Oh ya....... got it @shubhamsrg
sauravshakya
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So,from there we conclude that 5r<3q
shubhamsrg
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guess not,,how can you say that ?
sauravshakya
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Because if 5r>=3q then R.H.S will always be integer.
sauravshakya
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I mean L.H.S
shubhamsrg
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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..
sauravshakya
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oh wait........ I guess R.H.S can be integer too.
shubhamsrg
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also, we may always conclude that 7s > 2p.3q,5r ..but nothing can be said for inequalities in between 2p,3q and 5r..
sauravshakya
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Ok further simplification of it gives:
|dw:1349956922192:dw|
mukushla
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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\[2p=3q=5r\]\[7s-2p=1\]
sauravshakya
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GREAT.
sauravshakya
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Now, we need to find its minimum value.
mukushla
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what is that fraction?
sauravshakya
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p+q+r+s = (207p+15)/105
sauravshakya
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right?
mukushla
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ahh yeah :)
sauravshakya
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Now, we need to find its minimum integer value.
sauravshakya
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Oh wait...... But that can make other numbers decimal.
mukushla
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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
mukushla
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if u wanna check the answer i found is 106 for
\[m=6\]\[p=45\]\[q=30\]\[r=18\]\[s=13\]
sauravshakya
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Thanx @mukushla u r really amazing.