Question

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.

Answer 1

9^p+27^q+3^5r=3^7s

Answer 2

ok

Answer 3

got it @sauravshakya

Answer 4

What was that?

Answer 5

we just square and cube the value

Answer 6

The question is to find minimum value of p+q+r+s

Answer 7

then common 3 as a base

Answer 8

and then add powers

Answer 9

then you solve it

Answer 10

I have never seen that rule.

Answer 11

ok so go through wolframalpha

Answer 12

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

Answer 13

I dont think that will also help here.

Answer 14

go through

Answer 15

There is nothing.

Answer 16

I am seeing it.

Answer 17

So, what is the answer?

Answer 18

Give a min, but at tell u the answer can't be find the above way.

Answer 19

What I figured out till now is s must be an odd number....

Answer 20

@experimentX PLZ see this

Answer 21

this doesn't look nice ...
saying that find the minimum value of p+q+r+s <--- this should be close to first instances.

Answer 22

Cant we find that......

Answer 23

exponential diophantine equations are awful!! you should probably tag mukushla

Answer 24

This question is not of mine........ Someone posted this today and said a 8th grade challenging question.

Answer 25

But he is not online.

Answer 26

@mukushla

Answer 27

you don't do diophantine equations on 8th grade.

Answer 28

Even I dont know that.

Answer 29

This surely cant be a 8th grade question.

Answer 30

@mathslover i think he do this

Answer 31

very dhinchach question

Answer 32

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

Answer 33

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!!

Answer 34

I dont know it even has a solution.

Answer 35

*now ..
seeya later!!

Answer 36

ok...... BYE thanx for trying.

Answer 37

call it hardly a try .. hehehe

Answer 38

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 ) ??

Answer 39

@ujjwal p,q,r,s are only positive integers..

Answer 40

@shubhamsrg
( 3^(2p) + 3^(3q) + 3^(5r) )/3 >= 3^( (2p + 3q + 5r)/3 )
HOW???

Answer 41

from AM>=GM

Answer 42

Oh >= got it.

Answer 43

Yeah nice one.

Answer 44

and continuing that,
we have
21s >= 2p + 3q + 5r + 3

but still no idea if that'll help..

Answer 45

And I got: