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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.
 one year ago
 one year ago
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.
 one year ago
 one year ago

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mayankdevnaniBest ResponseYou've already chosen the best response.0
9^p+27^q+3^5r=3^7s
 one year ago

mayankdevnaniBest ResponseYou've already chosen the best response.0
got it @sauravshakya
 one year ago

mayankdevnaniBest ResponseYou've already chosen the best response.0
we just square and cube the value
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.0
The question is to find minimum value of p+q+r+s
 one year ago

mayankdevnaniBest ResponseYou've already chosen the best response.0
then common 3 as a base
 one year ago

mayankdevnaniBest ResponseYou've already chosen the best response.0
and then add powers
 one year ago

mayankdevnaniBest ResponseYou've already chosen the best response.0
then you solve it
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.0
I have never seen that rule.
 one year ago

mayankdevnaniBest ResponseYou've already chosen the best response.0
ok so go through wolframalpha
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.0
I dont think that will also help here.
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.0
There is nothing.
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.0
So, what is the answer?
 one year ago

ZekariasBest ResponseYou've already chosen the best response.0
Give a min, but at tell u the answer can't be find the above way.
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.0
What I figured out till now is s must be an odd number....
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.0
@experimentX PLZ see this
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
this doesn't look nice ... saying that find the minimum value of p+q+r+s < this should be close to first instances.
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.0
Cant we find that......
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
exponential diophantine equations are awful!! you should probably tag mukushla
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.0
This question is not of mine........ Someone posted this today and said a 8th grade challenging question.
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.0
But he is not online.
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
you don't do diophantine equations on 8th grade.
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.0
Even I dont know that.
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.0
This surely cant be a 8th grade question.
 one year ago

mayankdevnaniBest ResponseYou've already chosen the best response.0
@mathslover i think he do this
 one year ago

mayankdevnaniBest ResponseYou've already chosen the best response.0
very dhinchach question
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.0
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
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
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!!
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.0
I dont know it even has a solution.
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
*now .. seeya later!!
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.0
ok...... BYE thanx for trying.
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
call it hardly a try .. hehehe
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.0
how about this : i dont know if this'll work : ( 3^(2p) + 3^(3q) + 3^(5r) )/3 >= 3^( (2p + 3q + 5r)/3 ) or 3^(7s1) >= 3^( (2p + 3q + 5r)/3 ) ??
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.0
@ujjwal p,q,r,s are only positive integers..
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.0
@shubhamsrg ( 3^(2p) + 3^(3q) + 3^(5r) )/3 >= 3^( (2p + 3q + 5r)/3 ) HOW???
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.0
and continuing that, we have 21s >= 2p + 3q + 5r + 3 but still no idea if that'll help..
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.0
And I got: dw:1349955495301:dw
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.0
I guess NOW, we can prove it has no SOLUTION
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.0
THAT GIVES: dw:1349955687180:dw
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.0
Now, dw:1349955794330:dw
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.0
Now, can I say No solution...... Or I lack something.
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.0
well you have made an assumption that 5r<3q .. only that makes LHS an integer..otherwise you cant say..
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.0
am sorry,,you assumed 5r>3q ..
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.0
but conclusion to my saying is that you made an assumtion..
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.0
oh...... no assumtion.
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.0
L.H.S or R.H.S can be both negative
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.0
Just looking for integer....... L.H.S will always be integer R.H.S will always be in decimal.
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.0
LHS is not necessary an integer.. we only know that p,q,r,s are integers..
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.0
and agreed that RHS will always be noninteger..
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.0
Oh ya....... got it @shubhamsrg
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.0
So,from there we conclude that 5r<3q
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.0
guess not,,how can you say that ?
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.0
Because if 5r>=3q then R.H.S will always be integer.
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.0
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..
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.0
oh wait........ I guess R.H.S can be integer too.
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.0
also, we may always conclude that 7s > 2p.3q,5r ..but nothing can be said for inequalities in between 2p,3q and 5r..
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.0
Ok further simplification of it gives: dw:1349956922192:dw
 one year ago

mukushlaBest ResponseYou've already chosen the best response.6
divide both sides by \(3^{2p}\)\[1+3^{3q2p}+3^{5r2p}=3^{7s2p}\]right hand side is a power of 3 so does the left hand side so we must have\[2p=3q=5r\]\[7s2p=1\]
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.0
Now, we need to find its minimum value.
 one year ago

mukushlaBest ResponseYou've already chosen the best response.6
what is that fraction?
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.0
p+q+r+s = (207p+15)/105
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.0
Now, we need to find its minimum integer value.
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.0
Oh wait...... But that can make other numbers decimal.
 one year ago

mukushlaBest ResponseYou've already chosen the best response.6
u better start with \[7s2p=1\]ans\[p=\frac{7s1}{2}\]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
 one year ago

mukushlaBest ResponseYou've already chosen the best response.6
if u wanna check the answer i found is 106 for \[m=6\]\[p=45\]\[q=30\]\[r=18\]\[s=13\]
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.0
Thanx @mukushla u r really amazing.
 one year ago
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