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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.

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  1. mayankdevnani
    • 3 years ago
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    9^p+27^q+3^5r=3^7s

  2. mayankdevnani
    • 3 years ago
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    ok

  3. mayankdevnani
    • 3 years ago
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    got it @sauravshakya

  4. sauravshakya
    • 3 years ago
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    What was that?

  5. mayankdevnani
    • 3 years ago
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    we just square and cube the value

  6. sauravshakya
    • 3 years ago
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    The question is to find minimum value of p+q+r+s

  7. mayankdevnani
    • 3 years ago
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    then common 3 as a base

  8. mayankdevnani
    • 3 years ago
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    and then add powers

  9. mayankdevnani
    • 3 years ago
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    then you solve it

  10. sauravshakya
    • 3 years ago
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    I have never seen that rule.

  11. mayankdevnani
    • 3 years ago
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    ok so go through wolframalpha

  12. sauravshakya
    • 3 years ago
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    I dont think that will also help here.

  13. mayankdevnani
    • 3 years ago
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    go through

  14. sauravshakya
    • 3 years ago
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    There is nothing.

  15. Zekarias
    • 3 years ago
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    I am seeing it.

  16. sauravshakya
    • 3 years ago
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    So, what is the answer?

  17. Zekarias
    • 3 years ago
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    Give a min, but at tell u the answer can't be find the above way.

  18. sauravshakya
    • 3 years ago
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    What I figured out till now is s must be an odd number....

  19. sauravshakya
    • 3 years ago
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    @experimentX PLZ see this

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

  21. sauravshakya
    • 3 years ago
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    Cant we find that......

  22. experimentX
    • 3 years ago
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    exponential diophantine equations are awful!! you should probably tag mukushla

  23. sauravshakya
    • 3 years ago
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    This question is not of mine........ Someone posted this today and said a 8th grade challenging question.

  24. sauravshakya
    • 3 years ago
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    But he is not online.

  25. sauravshakya
    • 3 years ago
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    @mukushla

  26. experimentX
    • 3 years ago
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    you don't do diophantine equations on 8th grade.

  27. sauravshakya
    • 3 years ago
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    Even I dont know that.

  28. sauravshakya
    • 3 years ago
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    This surely cant be a 8th grade question.

  29. mayankdevnani
    • 3 years ago
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    @mathslover i think he do this

  30. mayankdevnani
    • 3 years ago
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    very dhinchach question

  31. sauravshakya
    • 3 years ago
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    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

  32. experimentX
    • 3 years ago
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    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!!

  33. sauravshakya
    • 3 years ago
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    I dont know it even has a solution.

  34. experimentX
    • 3 years ago
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    *now .. seeya later!!

  35. sauravshakya
    • 3 years ago
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    ok...... BYE thanx for trying.

  36. experimentX
    • 3 years ago
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    call it hardly a try .. hehehe

  37. shubhamsrg
    • 3 years ago
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    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 ) ??

  38. shubhamsrg
    • 3 years ago
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    @ujjwal p,q,r,s are only positive integers..

  39. sauravshakya
    • 3 years ago
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    @shubhamsrg ( 3^(2p) + 3^(3q) + 3^(5r) )/3 >= 3^( (2p + 3q + 5r)/3 ) HOW???

  40. shubhamsrg
    • 3 years ago
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    from AM>=GM

  41. sauravshakya
    • 3 years ago
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    Oh >= got it.

  42. sauravshakya
    • 3 years ago
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    Yeah nice one.

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

  44. sauravshakya
    • 3 years ago
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    And I got: |dw:1349955495301:dw|

  45. sauravshakya
    • 3 years ago
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    I guess NOW, we can prove it has no SOLUTION

  46. shubhamsrg
    • 3 years ago
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    hmm ??

  47. hartnn
    • 3 years ago
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    *

  48. sauravshakya
    • 3 years ago
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    THAT GIVES: |dw:1349955687180:dw|

  49. sauravshakya
    • 3 years ago
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    Now, |dw:1349955794330:dw|

  50. sauravshakya
    • 3 years ago
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    Now, can I say No solution...... Or I lack something.

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

  52. shubhamsrg
    • 3 years ago
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    am sorry,,you assumed 5r>3q ..

  53. shubhamsrg
    • 3 years ago
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    but conclusion to my saying is that you made an assumtion..

  54. sauravshakya
    • 3 years ago
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    oh...... no assumtion.

  55. sauravshakya
    • 3 years ago
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    L.H.S or R.H.S can be both negative

  56. sauravshakya
    • 3 years ago
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    Just looking for integer....... L.H.S will always be integer R.H.S will always be in decimal.

  57. shubhamsrg
    • 3 years ago
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    LHS is not necessary an integer.. we only know that p,q,r,s are integers..

  58. shubhamsrg
    • 3 years ago
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    and agreed that RHS will always be non-integer..

  59. sauravshakya
    • 3 years ago
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    Oh ya....... got it @shubhamsrg

  60. sauravshakya
    • 3 years ago
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    So,from there we conclude that 5r<3q

  61. shubhamsrg
    • 3 years ago
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    guess not,,how can you say that ?

  62. sauravshakya
    • 3 years ago
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    Because if 5r>=3q then R.H.S will always be integer.

  63. sauravshakya
    • 3 years ago
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    I mean L.H.S

  64. shubhamsrg
    • 3 years ago
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    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..

  65. sauravshakya
    • 3 years ago
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    oh wait........ I guess R.H.S can be integer too.

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

  67. sauravshakya
    • 3 years ago
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    Ok further simplification of it gives: |dw:1349956922192:dw|

  68. mukushla
    • 3 years ago
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    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\[2p=3q=5r\]\[7s-2p=1\]

  69. sauravshakya
    • 3 years ago
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    GREAT.

  70. sauravshakya
    • 3 years ago
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    Now, we need to find its minimum value.

  71. mukushla
    • 3 years ago
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    what is that fraction?

  72. sauravshakya
    • 3 years ago
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    p+q+r+s = (207p+15)/105

  73. sauravshakya
    • 3 years ago
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    right?

  74. mukushla
    • 3 years ago
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    ahh yeah :)

  75. sauravshakya
    • 3 years ago
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    Now, we need to find its minimum integer value.

  76. sauravshakya
    • 3 years ago
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    Oh wait...... But that can make other numbers decimal.

  77. mukushla
    • 3 years ago
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    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

  78. mukushla
    • 3 years ago
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    if u wanna check the answer i found is 106 for \[m=6\]\[p=45\]\[q=30\]\[r=18\]\[s=13\]

  79. sauravshakya
    • 3 years ago
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    Thanx @mukushla u r really amazing.

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