Hello,
x,y,z∈ IN
[tex] x=\frac{3}{4}*y [/tex]
[tex] y=\frac{1}{3}*z [/tex]
[tex]140<z<151\ ==>\ \frac{140}{3}<\frac{z}{3}<\frac{151}{3}[/tex]
[tex]46\leq\ y\ \leq\ 50 ==>\ \frac{46*3}{4}\leq\frac{3y}{4}\leq\frac{50*3}{4}[/tex]
[tex]35\ \leq\ z\ \leq\ 37[/tex]==>
x∈{35,36,37}
y∈{47,48,49}
z∈{141,144,147}
We must go futher:
if x=35 then y=4*35/3=46.666 not good , not whole
if x=36 then y=4*36/3=48 ok
if x=37 then y=37*4/3=49.333.. not good
So there is only one solution (36,48,144) for (x,y,z)
No problem with z for z=3/1*4/3*x=4x ==>whole.
