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anonymous
 3 years 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.
anonymous
 3 years 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.

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mayankdevnani
 3 years ago
Best ResponseYou've already chosen the best response.09^p+27^q+3^5r=3^7s

mayankdevnani
 3 years ago
Best ResponseYou've already chosen the best response.0got it @sauravshakya

mayankdevnani
 3 years ago
Best ResponseYou've already chosen the best response.0we just square and cube the value

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The question is to find minimum value of p+q+r+s

mayankdevnani
 3 years ago
Best ResponseYou've already chosen the best response.0then common 3 as a base

mayankdevnani
 3 years ago
Best ResponseYou've already chosen the best response.0and then add powers

mayankdevnani
 3 years ago
Best ResponseYou've already chosen the best response.0then you solve it

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I have never seen that rule.

mayankdevnani
 3 years ago
Best ResponseYou've already chosen the best response.0ok so go through wolframalpha

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I dont think that will also help here.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So, what is the answer?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Give a min, but at tell u the answer can't be find the above way.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0What I figured out till now is s must be an odd number....

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@experimentX PLZ see this

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Cant we find that......

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0exponential diophantine equations are awful!! you should probably tag mukushla

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0But he is not online.

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0you don't do diophantine equations on 8th grade.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Even I dont know that.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0This surely cant be a 8th grade question.

mayankdevnani
 3 years ago
Best ResponseYou've already chosen the best response.0@mathslover i think he do this

mayankdevnani
 3 years ago
Best ResponseYou've already chosen the best response.0very dhinchach question

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Now, 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
 3 years ago
Best ResponseYou've already chosen the best response.0anyway 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
 3 years ago
Best ResponseYou've already chosen the best response.0I dont know it even has a solution.

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0*now .. seeya later!!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok...... BYE thanx for trying.

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0call it hardly a try .. hehehe

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.0how 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 ) ??

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.0@ujjwal p,q,r,s are only positive integers..

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0And I got: dw:1349955495301:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I guess NOW, we can prove it has no SOLUTION

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0THAT GIVES: dw:1349955687180:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Now, dw:1349955794330:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Now, can I say No solution...... Or I lack something.

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

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.0am sorry,,you assumed 5r>3q ..

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.0but conclusion to my saying is that you made an assumtion..

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh...... no assumtion.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0L.H.S or R.H.S can be both negative

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Just looking for integer....... L.H.S will always be integer R.H.S will always be in decimal.

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

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.0and agreed that RHS will always be noninteger..

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Oh ya....... got it @shubhamsrg

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So,from there we conclude that 5r<3q

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.0guess not,,how can you say that ?

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

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.0but 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
 3 years ago
Best ResponseYou've already chosen the best response.0oh wait........ I guess R.H.S can be integer too.

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Ok further simplification of it gives: dw:1349956922192:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0divide 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\[2p=3q=5r\]\[7s2p=1\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Now, we need to find its minimum value.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0what is that fraction?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0p+q+r+s = (207p+15)/105

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Now, we need to find its minimum integer value.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Oh wait...... But that can make other numbers decimal.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0u 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0if u wanna check the answer i found is 106 for \[m=6\]\[p=45\]\[q=30\]\[r=18\]\[s=13\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Thanx @mukushla u r really amazing.
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