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sauravshakya

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.

  • one year ago
  • one year ago

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

    • one year ago
  2. mayankdevnani
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    ok

    • one year ago
  3. mayankdevnani
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    got it @sauravshakya

    • one year ago
  4. sauravshakya
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    What was that?

    • one year ago
  5. mayankdevnani
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    we just square and cube the value

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

    • one year ago
  7. mayankdevnani
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    then common 3 as a base

    • one year ago
  8. mayankdevnani
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    and then add powers

    • one year ago
  9. mayankdevnani
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    then you solve it

    • one year ago
  10. sauravshakya
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    I have never seen that rule.

    • one year ago
  11. mayankdevnani
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    ok so go through wolframalpha

    • one year ago
  12. sauravshakya
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    I dont think that will also help here.

    • one year ago
  13. mayankdevnani
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    go through

    • one year ago
  14. sauravshakya
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    There is nothing.

    • one year ago
  15. Zekarias
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    I am seeing it.

    • one year ago
  16. sauravshakya
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    So, what is the answer?

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

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

    • one year ago
  19. sauravshakya
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    @experimentX PLZ see this

    • one year ago
  20. experimentX
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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.

    • one year ago
  21. sauravshakya
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    Cant we find that......

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

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

    • one year ago
  24. sauravshakya
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    But he is not online.

    • one year ago
  25. sauravshakya
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    @mukushla

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

    • one year ago
  27. sauravshakya
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    Even I dont know that.

    • one year ago
  28. sauravshakya
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    This surely cant be a 8th grade question.

    • one year ago
  29. mayankdevnani
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    @mathslover i think he do this

    • one year ago
  30. mayankdevnani
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    very dhinchach question

    • one year ago
  31. sauravshakya
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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

    • one year ago
  32. experimentX
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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!!

    • one year ago
  33. sauravshakya
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    I dont know it even has a solution.

    • one year ago
  34. experimentX
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    *now .. seeya later!!

    • one year ago
  35. sauravshakya
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    ok...... BYE thanx for trying.

    • one year ago
  36. experimentX
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    call it hardly a try .. hehehe

    • one year ago
  37. shubhamsrg
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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 ) ??

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

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

    • one year ago
  40. shubhamsrg
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    from AM>=GM

    • one year ago
  41. sauravshakya
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    Oh >= got it.

    • one year ago
  42. sauravshakya
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    Yeah nice one.

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

    • one year ago
  44. sauravshakya
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    And I got: |dw:1349955495301:dw|

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

    • one year ago
  46. shubhamsrg
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    hmm ??

    • one year ago
  47. hartnn
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    *

    • one year ago
  48. sauravshakya
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    THAT GIVES: |dw:1349955687180:dw|

    • one year ago
  49. sauravshakya
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    Now, |dw:1349955794330:dw|

    • one year ago
  50. sauravshakya
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    Now, can I say No solution...... Or I lack something.

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

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

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

    • one year ago
  54. sauravshakya
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    oh...... no assumtion.

    • one year ago
  55. sauravshakya
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    L.H.S or R.H.S can be both negative

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

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

    • one year ago
  58. shubhamsrg
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    and agreed that RHS will always be non-integer..

    • one year ago
  59. sauravshakya
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    Oh ya....... got it @shubhamsrg

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

    • one year ago
  61. shubhamsrg
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    guess not,,how can you say that ?

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

    • one year ago
  63. sauravshakya
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    I mean L.H.S

    • one year ago
  64. shubhamsrg
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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..

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

    • one year ago
  66. shubhamsrg
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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..

    • one year ago
  67. sauravshakya
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    Ok further simplification of it gives: |dw:1349956922192:dw|

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

    • one year ago
  69. sauravshakya
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    GREAT.

    • one year ago
  70. sauravshakya
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    Now, we need to find its minimum value.

    • one year ago
  71. mukushla
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    what is that fraction?

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

    • one year ago
  73. sauravshakya
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    right?

    • one year ago
  74. mukushla
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    ahh yeah :)

    • one year ago
  75. sauravshakya
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    Now, we need to find its minimum integer value.

    • one year ago
  76. sauravshakya
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    Oh wait...... But that can make other numbers decimal.

    • one year ago
  77. mukushla
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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

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

    • one year ago
  79. sauravshakya
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    Thanx @mukushla u r really amazing.

    • one year ago
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