NMTC Practice Part 9

$97 \times (yz-z-1)=19xyz$ $\Rightarrow 19xyz \lt 97yz \Rightarrow x \lt 6$ It is easy to get: x=5, y=49, x=97. $Res=\boxed{151}$

Sorry for not formatting. I hope this will do

1/x-1/xy-1/xyz=19/97,x<y<z,x,y,z∈N

x+y+z= ?

1/6<19/97<1/5. Since (-1/xy-1/xyz)<0, x≤5.

1/x-1/xy-1/xyz=19/97

yz/xyz-z/xyz-1/xyz=19/97

(yz-z-1)/xyz=19/97

97yz-97z-97=19xyz

19xyz-97yz+97z=-97

z(19xy-97y+97)=-97

97=(-1)(97), or -97=(1)(-97). z≠-1,-97 because z∈N. . z≠1 because x<y<z.

If z=1, then x<y<1, which would contradict x,y∈N. Therefore...

(z=97)

97(19xy-97y+97)=-97

19xy-97y+97=-1

97y-19xy=98

y(97-19x)=98

98=(2)(7)(7). This leaves 4 possible solutions for 97-19x

97-19x=2,7,14,49

-19x=-95,-90,-83,-48,

x=5,90/19,83/19,48/19 Since 5 is the only natural number,

(x=5)

y(97-19(5))=98

y(97-95)=98

2y=98

(y=49)

These solutions satisfy all three criteria.

5<49<97

5,49,97∈N

1/5-1/(5)(49) -1/(5)(49)(97) =19/97

(49)(97)/(5)(49)(97) -97/(5)(49)(97) -1/(5)(49)(97) =19/97

4655/23765=19/97

19/97=19/97

(x+y+z=151)

claim 1: x<=5. proof: 1/x=19/97+1/xy+1/xyz > 19/97 This implies that: x<=5<97/19.

claim 2:(a)z>=3 {proof is very simple} (b) z=97. proof: upon simplifying the expression we get that; (y*z-z-1)/xyz=19/97, here, z does not divide the numerator and hence is the factor of 97 which is a prime. thus, z=97.

now, plug in z=97, and upon simplification the expression becomes; (97-19x)y=98=7 7 2=(2+(5-x)19)y, now check for values of x from 5 to 1, and you 'll get the desired conclusion.