Question

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

Answer 1

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

Answer 2

correction
\[\frac{ 97 }{ 19 yz }\left( z+1 \right)\]

Answer 3

what about y and z then

Answer 4

\[x=\frac{97yz-97z-97}{19yz} \implies 19xyz = 97(yz-z-1) \\~\\~\\ \implies z =\dfrac{97}{97y-19xy-97}\]

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 \(97-19x \in \{\text{positive factors of 98}\}\)

Answer 5

yz-z-1>0
yz>z+1
yz>z
yz-z>0
(y-1)z>0
y>1
so \[y \ge2,z>y \ge2,z \ge3\]

Answer 6

I'm confused... how do I connect all of these equations

Answer 7

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.

Answer 8

Should I put 97 into the equation now?

Answer 9

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

Answer 10

the problem says there is an integral value of y that gives positive integral value of z

Answer 11

in the statement it is also written x,y,z are positive integers

Answer 12

If no integral value of y gives positive integral value of z then what o.o

Answer 13

do you find anything wrong in my calculation?

Answer 14

not really, no

Answer 15

Well thanks, I guess