A community for students. Sign up today!
Here's the question you clicked on:
 0 viewing
 2 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.
 2 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.

This Question is Closed

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

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

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

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

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

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

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

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

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

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

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

Zekarias
 2 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.

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

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

experimentX
 2 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.

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

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

sauravshakya
 2 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.

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

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

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

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

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

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

sauravshakya
 2 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
 2 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!!

sauravshakya
 2 years ago
Best ResponseYou've already chosen the best response.0I dont know it even has a solution.

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

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

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

shubhamsrg
 2 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
 2 years ago
Best ResponseYou've already chosen the best response.0@ujjwal p,q,r,s are only positive integers..

sauravshakya
 2 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
 2 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..

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

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

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

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

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

shubhamsrg
 2 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
 2 years ago
Best ResponseYou've already chosen the best response.0am sorry,,you assumed 5r>3q ..

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

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

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

sauravshakya
 2 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
 2 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
 2 years ago
Best ResponseYou've already chosen the best response.0and agreed that RHS will always be noninteger..

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

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

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

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

shubhamsrg
 2 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..

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

shubhamsrg
 2 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..

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

mukushla
 2 years ago
Best ResponseYou've already chosen the best response.6divide 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\[2p=3q=5r\]\[7s2p=1\]

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

mukushla
 2 years ago
Best ResponseYou've already chosen the best response.6what is that fraction?

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

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

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

mukushla
 2 years ago
Best ResponseYou've already chosen the best response.6u 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

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

sauravshakya
 2 years ago
Best ResponseYou've already chosen the best response.0Thanx @mukushla u r really amazing.
Ask your own question
Ask a QuestionFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.