Question:- The sum of three positive integers is 396 and their HCF is 12. The LCM of the smallest and the largest numbers is 1224. The LCM of the largest two numbers is 2040. Find the difference between the smallest and the largest number.
My approach:-
I took the three numbers to be $A=12*x$, $B=12*y$, $C=12*z$ where x, y, z are co-prime to each other and $x<y<z$
$12*(x+y+z)=396$
$x+y+z=33 -----(1)$
Now, LCM of largest and smallest number would be $12*x*z=1224$ and LCM of the largest two numbers would be $12*y*z=2040$ ;
On dividing these 2 equations, we would get :-
$y/x = 5/3$ , hence $x=3k$ and $y=5k$ , k has to 1 as otherwise x and y wont be co-prime as per our initial condition
hence $x=3, y=5$ , putting this in the Equation (1),
$3+5+z=33$
$z=25$ , but now z is not anymore coprime with y
Where am I going wrong with my steps ?
Edit:
After reading the comments given, I have changed my solution as below, please let me know if it is correct now ?
$12*LCM(x,z)=1224$ -> $LCM(x,z)=102$
$102 = 2*3*17$
Similarly, $LCM(y,z)=170$
$170=2*5*17$
y will have $5^1$ in its prime factorization and x will have $3^1$ in prime factorization and now $17^1$ will need to be present for z as otherwise x and y will be exceeding 33
$x=2^a * 3$ (possible values of x will be 3,6,24)
$z=17$
$y=2^q * 5$ (possible values of y will be 5,10,20)
and now using the the above equations in (1)
$x+y+17=33$
$x+y=16$
$x=6$ and $y=10$ is the only possible solution adhering to the possible solutions we listed previously therefore the final 3 numbers will be
$12*6 , 12*10 , 12*17$
$72, 120, 204$
and my answer should be $204-72 = 132$