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h0pe
 one year ago
Find the unique triple \(x,y,z\) of positive integers such that \(x < y < z\) and
\[x=\frac{97yz97z97}{19yz}\].
h0pe
 one year ago
Find the unique triple \(x,y,z\) of positive integers such that \(x < y < z\) and \[x=\frac{97yz97z97}{19yz}\].

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[x=\frac{ 97yz97z97 }{ 19yz }=\frac{ 97 }{ 19 }\frac{ 97 }{ 19 yz }\left( y+1 \right)\] so \[x<\frac{ 97 }{ 19 }\] but x is a positive integer. so x can be 1,2,3,4,5

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0correction \[\frac{ 97 }{ 19 yz }\left( z+1 \right)\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0\[x=\frac{97yz97z97}{19yz} \implies 19xyz = 97(yzz1) \\~\\~\\ \implies z =\dfrac{97}{97y19xy97}\] Since \(z\gt 0\) and \(z\gt y\gt x\), it must be the case that \[97y19xy97 = 1\] isolating \(y\) : \[y = \dfrac {98}{9719x}\] so \(9719x \in \{\text{positive factors of 98}\}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yzz1>0 yz>z+1 yz>z yzz>0 (y1)z>0 y>1 so \[y \ge2,z>y \ge2,z \ge3\]

h0pe
 one year ago
Best ResponseYou've already chosen the best response.0I'm confused... how do I connect all of these equations

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well, consider what ganeshie had. He had it worked down to y = 98/(9719x), which means 9719x 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. 9719x = 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.

h0pe
 one year ago
Best ResponseYou've already chosen the best response.0Should I put 97 into the equation now?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0if 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 yz97z97 \[40yz97 z=97\] \[z=\frac{ 97 }{ 40 y97 }\] no integral value of y gives positive integral value of z

h0pe
 one year ago
Best ResponseYou've already chosen the best response.0the problem says there is an integral value of y that gives positive integral value of z

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0in the statement it is also written x,y,z are positive integers

h0pe
 one year ago
Best ResponseYou've already chosen the best response.0If no integral value of y gives positive integral value of z then what o.o

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0do you find anything wrong in my calculation?
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