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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 $204-72 = 132$

Fin27
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    You don't need $x$ and $y$ to be coprime (i.e. $\gcd(x,y)=1$) you just need $\gcd(x,y,z)=1$ – Zoe Allen Mar 18 '23 at 14:12
  • So are you saying that at the step $y/x = 5/3$ , I should take x=3k and 5k and not just one case of x=3 and y=5 ? Is the initial part of the solution where I am taking the LCM fine and if its wrong then how shall I form the LCM equations then? – Fin27 Mar 18 '23 at 20:05
  • No, there is also an error in the initial part where you seem to be assume (w/o proof) that $x$ and $y$ are both coprime to $z,,$ so have lcm = product. Correct is ${\rm lcm}(A,C)={\rm lcm}(12x,12z) = 12,{\rm lcm}(x,z),$ which $= 12xz,$ only when $x,z$ are coprime. – Bill Dubuque Mar 18 '23 at 20:14
  • Thank you I understand that I can't write 12xz as LCM since x and z need not be co-prime, in that case how shall I utilize the 2 LCM equations given, because I feel that they will be needed to solve the question further – Fin27 Mar 18 '23 at 20:18
  • $\begin{align} {\rm lcm}(x,z) = 2\cdot \color{#c00}3\cdot 17 &,\Rightarrow, 3\mid x\ ,{\rm or},\ \color{#c00}{3\mid y}\ {\rm lcm}(y,z)=2\cdot 5\cdot 17 &,\Rightarrow, 3\nmid z:\ &:\ \color{#c00}{3\nmid y} \end{align}\ \Rightarrow\ 3\mid x;\ {\rm Similarly}\ 5\mid y.,$ Further $z = 17,,$ else sum $>33$ – Bill Dubuque Mar 18 '23 at 20:53
  • I too have edited my solution can you please check it – Fin27 Mar 18 '23 at 20:57
  • Thank you Bill and Zoe :) – Fin27 Mar 18 '23 at 21:00

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