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sauravshakya
Group Title
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
 2 years ago
sauravshakya Group Title
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
 2 years ago

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mayankdevnani Group TitleBest ResponseYou've already chosen the best response.0
9^p+27^q+3^5r=3^7s
 2 years ago

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

sauravshakya Group TitleBest ResponseYou've already chosen the best response.0
What was that?
 2 years ago

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

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

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

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

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

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

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

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

mayankdevnani Group TitleBest ResponseYou've already chosen the best response.0
go through
 2 years ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.0
There is nothing.
 2 years ago

Zekarias Group TitleBest ResponseYou've already chosen the best response.0
I am seeing it.
 2 years ago

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

Zekarias Group TitleBest ResponseYou've already chosen the best response.0
Give a min, but at tell u the answer can't be find the above way.
 2 years ago

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

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

experimentX Group TitleBest 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.
 2 years ago

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

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

sauravshakya Group TitleBest 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.
 2 years ago

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

sauravshakya Group TitleBest ResponseYou've already chosen the best response.0
@mukushla
 2 years ago

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

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

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

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

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

sauravshakya Group TitleBest 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
 2 years ago

experimentX Group TitleBest 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!!
 2 years ago

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

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

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

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

shubhamsrg Group TitleBest 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 ) ??
 2 years ago

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

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

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
from AM>=GM
 2 years ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.0
Oh >= got it.
 2 years ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.0
Yeah nice one.
 2 years ago

shubhamsrg Group TitleBest 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..
 2 years ago

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

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

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

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

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

shubhamsrg Group TitleBest 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..
 2 years ago

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

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

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

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

sauravshakya Group TitleBest 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.
 2 years ago

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

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

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

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

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

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

sauravshakya Group TitleBest ResponseYou've already chosen the best response.0
I mean L.H.S
 2 years ago

shubhamsrg Group TitleBest 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..
 2 years ago

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

shubhamsrg Group TitleBest 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..
 2 years ago

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

mukushla Group TitleBest 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\[2p=3q=5r\]\[7s2p=1\]
 2 years ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.0
GREAT.
 2 years ago

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

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

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

sauravshakya Group TitleBest ResponseYou've already chosen the best response.0
right?
 2 years ago

mukushla Group TitleBest ResponseYou've already chosen the best response.6
ahh yeah :)
 2 years ago

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

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

mukushla Group TitleBest 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
 2 years ago

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

sauravshakya Group TitleBest ResponseYou've already chosen the best response.0
Thanx @mukushla u r really amazing.
 2 years ago
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