anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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mayankdevnani
  • mayankdevnani
9^p+27^q+3^5r=3^7s
mayankdevnani
  • mayankdevnani
ok
mayankdevnani
  • mayankdevnani
got it @sauravshakya

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More answers

anonymous
  • anonymous
What was that?
mayankdevnani
  • mayankdevnani
we just square and cube the value
anonymous
  • anonymous
The question is to find minimum value of p+q+r+s
mayankdevnani
  • mayankdevnani
then common 3 as a base
mayankdevnani
  • mayankdevnani
and then add powers
mayankdevnani
  • mayankdevnani
then you solve it
anonymous
  • anonymous
I have never seen that rule.
mayankdevnani
  • mayankdevnani
ok so go through wolframalpha
mayankdevnani
  • mayankdevnani
http://www.wolframalpha.com/input/?i=3%5E%282p%29+%2B+3%5E%283q%29+%2B+3%5E%285r%29+%3D+3%5E%287s%29+Find+the+minimum+value+of+p%2Bq%2Br%2Bs
anonymous
  • anonymous
I dont think that will also help here.
mayankdevnani
  • mayankdevnani
go through
anonymous
  • anonymous
There is nothing.
anonymous
  • anonymous
I am seeing it.
anonymous
  • anonymous
So, what is the answer?
anonymous
  • anonymous
Give a min, but at tell u the answer can't be find the above way.
anonymous
  • anonymous
What I figured out till now is s must be an odd number....
anonymous
  • anonymous
@experimentX PLZ see this
experimentX
  • experimentX
this doesn't look nice ... saying that find the minimum value of p+q+r+s <--- this should be close to first instances.
anonymous
  • anonymous
Cant we find that......
experimentX
  • experimentX
exponential diophantine equations are awful!! you should probably tag mukushla
anonymous
  • anonymous
This question is not of mine........ Someone posted this today and said a 8th grade challenging question.
anonymous
  • anonymous
But he is not online.
anonymous
  • anonymous
@mukushla
experimentX
  • experimentX
you don't do diophantine equations on 8th grade.
anonymous
  • anonymous
Even I dont know that.
anonymous
  • anonymous
This surely cant be a 8th grade question.
mayankdevnani
  • mayankdevnani
@mathslover i think he do this
mayankdevnani
  • mayankdevnani
very dhinchach question
anonymous
  • anonymous
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
  • experimentX
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!!
anonymous
  • anonymous
I dont know it even has a solution.
experimentX
  • experimentX
*now .. seeya later!!
anonymous
  • anonymous
ok...... BYE thanx for trying.
experimentX
  • experimentX
call it hardly a try .. hehehe
shubhamsrg
  • shubhamsrg
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
  • shubhamsrg
@ujjwal p,q,r,s are only positive integers..
anonymous
  • anonymous
@shubhamsrg ( 3^(2p) + 3^(3q) + 3^(5r) )/3 >= 3^( (2p + 3q + 5r)/3 ) HOW???
shubhamsrg
  • shubhamsrg
from AM>=GM
anonymous
  • anonymous
Oh >= got it.
anonymous
  • anonymous
Yeah nice one.
shubhamsrg
  • shubhamsrg
and continuing that, we have 21s >= 2p + 3q + 5r + 3 but still no idea if that'll help..
anonymous
  • anonymous
And I got: |dw:1349955495301:dw|
anonymous
  • anonymous
I guess NOW, we can prove it has no SOLUTION
shubhamsrg
  • shubhamsrg
hmm ??
hartnn
  • hartnn
*
anonymous
  • anonymous
THAT GIVES: |dw:1349955687180:dw|
anonymous
  • anonymous
Now, |dw:1349955794330:dw|
anonymous
  • anonymous
Now, can I say No solution...... Or I lack something.
shubhamsrg
  • shubhamsrg
well you have made an assumption that 5r<3q .. only that makes LHS an integer..otherwise you cant say..
shubhamsrg
  • shubhamsrg
am sorry,,you assumed 5r>3q ..
shubhamsrg
  • shubhamsrg
but conclusion to my saying is that you made an assumtion..
anonymous
  • anonymous
oh...... no assumtion.
anonymous
  • anonymous
L.H.S or R.H.S can be both negative
anonymous
  • anonymous
Just looking for integer....... L.H.S will always be integer R.H.S will always be in decimal.
shubhamsrg
  • shubhamsrg
LHS is not necessary an integer.. we only know that p,q,r,s are integers..
shubhamsrg
  • shubhamsrg
and agreed that RHS will always be non-integer..
anonymous
  • anonymous
Oh ya....... got it @shubhamsrg
anonymous
  • anonymous
So,from there we conclude that 5r<3q
shubhamsrg
  • shubhamsrg
guess not,,how can you say that ?
anonymous
  • anonymous
Because if 5r>=3q then R.H.S will always be integer.
anonymous
  • anonymous
I mean L.H.S
shubhamsrg
  • shubhamsrg
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..
anonymous
  • anonymous
oh wait........ I guess R.H.S can be integer too.
shubhamsrg
  • shubhamsrg
also, we may always conclude that 7s > 2p.3q,5r ..but nothing can be said for inequalities in between 2p,3q and 5r..
anonymous
  • anonymous
Ok further simplification of it gives: |dw:1349956922192:dw|
anonymous
  • anonymous
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\[2p=3q=5r\]\[7s-2p=1\]
anonymous
  • anonymous
GREAT.
anonymous
  • anonymous
Now, we need to find its minimum value.
anonymous
  • anonymous
what is that fraction?
anonymous
  • anonymous
p+q+r+s = (207p+15)/105
anonymous
  • anonymous
right?
anonymous
  • anonymous
ahh yeah :)
anonymous
  • anonymous
Now, we need to find its minimum integer value.
anonymous
  • anonymous
Oh wait...... But that can make other numbers decimal.
anonymous
  • anonymous
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
anonymous
  • anonymous
if u wanna check the answer i found is 106 for \[m=6\]\[p=45\]\[q=30\]\[r=18\]\[s=13\]
anonymous
  • anonymous
Thanx @mukushla u r really amazing.

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