## sauravshakya 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.

1. mayankdevnani

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

2. mayankdevnani

ok

3. mayankdevnani

got it @sauravshakya

4. sauravshakya

What was that?

5. mayankdevnani

we just square and cube the value

6. sauravshakya

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

7. mayankdevnani

then common 3 as a base

8. mayankdevnani

9. mayankdevnani

then you solve it

10. sauravshakya

I have never seen that rule.

11. mayankdevnani

ok so go through wolframalpha

12. mayankdevnani
13. sauravshakya

I dont think that will also help here.

14. mayankdevnani

go through

15. sauravshakya

There is nothing.

16. Zekarias

I am seeing it.

17. sauravshakya

18. Zekarias

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

19. sauravshakya

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

20. sauravshakya

@experimentX PLZ see this

21. experimentX

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

22. sauravshakya

Cant we find that......

23. experimentX

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

24. sauravshakya

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

25. sauravshakya

But he is not online.

26. sauravshakya

@mukushla

27. experimentX

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

28. sauravshakya

Even I dont know that.

29. sauravshakya

This surely cant be a 8th grade question.

30. mayankdevnani

@mathslover i think he do this

31. mayankdevnani

very dhinchach question

32. sauravshakya

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

33. 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!!

34. sauravshakya

I dont know it even has a solution.

35. experimentX

*now .. seeya later!!

36. sauravshakya

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

37. experimentX

call it hardly a try .. hehehe

38. 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 ) ??

39. shubhamsrg

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

40. sauravshakya

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

41. shubhamsrg

from AM>=GM

42. sauravshakya

Oh >= got it.

43. sauravshakya

Yeah nice one.

44. shubhamsrg

and continuing that, we have 21s >= 2p + 3q + 5r + 3 but still no idea if that'll help..

45. sauravshakya

And I got: |dw:1349955495301:dw|

46. sauravshakya

I guess NOW, we can prove it has no SOLUTION

47. shubhamsrg

hmm ??

48. hartnn

*

49. sauravshakya

THAT GIVES: |dw:1349955687180:dw|

50. sauravshakya

Now, |dw:1349955794330:dw|

51. sauravshakya

Now, can I say No solution...... Or I lack something.

52. shubhamsrg

well you have made an assumption that 5r<3q .. only that makes LHS an integer..otherwise you cant say..

53. shubhamsrg

am sorry,,you assumed 5r>3q ..

54. shubhamsrg

but conclusion to my saying is that you made an assumtion..

55. sauravshakya

oh...... no assumtion.

56. sauravshakya

L.H.S or R.H.S can be both negative

57. sauravshakya

Just looking for integer....... L.H.S will always be integer R.H.S will always be in decimal.

58. shubhamsrg

LHS is not necessary an integer.. we only know that p,q,r,s are integers..

59. shubhamsrg

and agreed that RHS will always be non-integer..

60. sauravshakya

Oh ya....... got it @shubhamsrg

61. sauravshakya

So,from there we conclude that 5r<3q

62. shubhamsrg

guess not,,how can you say that ?

63. sauravshakya

Because if 5r>=3q then R.H.S will always be integer.

64. sauravshakya

I mean L.H.S

65. 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..

66. sauravshakya

oh wait........ I guess R.H.S can be integer too.

67. shubhamsrg

also, we may always conclude that 7s > 2p.3q,5r ..but nothing can be said for inequalities in between 2p,3q and 5r..

68. sauravshakya

Ok further simplification of it gives: |dw:1349956922192:dw|

69. mukushla

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$

70. sauravshakya

GREAT.

71. sauravshakya

Now, we need to find its minimum value.

72. mukushla

what is that fraction?

73. sauravshakya

p+q+r+s = (207p+15)/105

74. sauravshakya

right?

75. mukushla

ahh yeah :)

76. sauravshakya

Now, we need to find its minimum integer value.

77. sauravshakya

Oh wait...... But that can make other numbers decimal.

78. mukushla

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

79. mukushla

if u wanna check the answer i found is 106 for $m=6$$p=45$$q=30$$r=18$$s=13$

80. sauravshakya

Thanx @mukushla u r really amazing.

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