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h0pe

  • one year ago

Find the unique triple \(x,y,z\) of positive integers such that \(x < y < z\) and \[x=\frac{97yz-97z-97}{19yz}\].

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  1. anonymous
    • one year ago
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    \[x=\frac{ 97yz-97z-97 }{ 19yz }=\frac{ 97 }{ 19 }-\frac{ 97 }{ 19 yz }\left( y+1 \right)\] so \[x<\frac{ 97 }{ 19 }\] but x is a positive integer. so x can be 1,2,3,4,5

  2. anonymous
    • one year ago
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    correction \[\frac{ 97 }{ 19 yz }\left( z+1 \right)\]

  3. h0pe
    • one year ago
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    what about y and z then

  4. ganeshie8
    • one year ago
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    \[x=\frac{97yz-97z-97}{19yz} \implies 19xyz = 97(yz-z-1) \\~\\~\\ \implies z =\dfrac{97}{97y-19xy-97}\] Since \(z\gt 0\) and \(z\gt y\gt x\), it must be the case that \[97y-19xy-97 = 1\] isolating \(y\) : \[y = \dfrac {98}{97-19x}\] so \(97-19x \in \{\text{positive factors of 98}\}\)

  5. anonymous
    • one year ago
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    yz-z-1>0 yz>z+1 yz>z yz-z>0 (y-1)z>0 y>1 so \[y \ge2,z>y \ge2,z \ge3\]

  6. h0pe
    • one year ago
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    I'm confused... how do I connect all of these equations

  7. anonymous
    • one year ago
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    Well, consider what ganeshie had. He had it worked down to y = 98/(97-19x), which means 97-19x had to be equal to a factor of 98. Factors of 98 are 2, 7, and 49. The only factor of 98 that makes the value of x equal to a positive integer is 2. 97-19x = 2 -19x = -95 x = 5 The other possible factors of 98 would fail. So since we can confirm x is 5, we would also get y to be 98/2 = 49 And from before \[z = \frac{ 97 }{ 97y -19xy -1 }\] We confirmed that 97y -19xy -1 had to be equal to 1, which means z is 97.

  8. h0pe
    • one year ago
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    Should I put 97 into the equation now?

  9. anonymous
    • one year ago
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    if we try one by one no value of satisfy this . Hence no positive integers are possible with the given condition. Let us take the case x=3 57 yz=97 yz-97z-97 \[40yz-97 z=97\] \[z=\frac{ 97 }{ 40 y-97 }\] no integral value of y gives positive integral value of z

  10. h0pe
    • one year ago
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    the problem says there is an integral value of y that gives positive integral value of z

  11. anonymous
    • one year ago
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    in the statement it is also written x,y,z are positive integers

  12. h0pe
    • one year ago
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    If no integral value of y gives positive integral value of z then what o.o

  13. anonymous
    • one year ago
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    do you find anything wrong in my calculation?

  14. h0pe
    • one year ago
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    not really, no

  15. h0pe
    • one year ago
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    Well thanks, I guess

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