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Given a system:

$$ \begin{cases} 5x - 3y = 0 \\ \text{lcm}(x, y) = 45 \end{cases} $$

Since, $5x=3y$, I've tried expressing product of $xy$ as:

$$ \text{lcm}(x,y)\gcd(x,y)=xy \\ \gcd(x,y)=\frac{xy}{\text{lcm}(x,y)} $$

I see that $x = 9$ and $y = 15$, I understand that we get it from dividing $\text{lcm}(x,y)$ which is 45 by coefficients, but I don't understand why.

Bill Dubuque
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Felix
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  • Well, there aren't very many pairs $(x,y)$ with $\text {lcm}(x,y)=45$, you could just try all of them. And you know $3,|,x$ and $5,|,y$ which narrows the list further. – lulu Sep 17 '23 at 23:41
  • By the linked LCM Distributive Law we can scale by $,\color{#c00}3,$ so we can use $,\color{#c00}{3y=5x},$ to eliminate $,y,$ yielding $$3\cdot 45 = 3,{\rm lcm}(x,y) = {\rm lcm}(3x,\color{#c00}{3y}) = {\rm lcm}(3x,\color{#c00}{5x}) = {\rm lcm}(3,5)x = 15x\ \Rightarrow\ x = 9\qquad$$ – Bill Dubuque Sep 18 '23 at 00:03
  • Alternatively we can work with prime factorizations of $,x,y\mid {\rm lcm}(x,y)=45,$ as here or here or the answer below, or many other prior answers. – Bill Dubuque Sep 18 '23 at 02:59

1 Answers1

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If $\mathrm{lcm}(x, y) = 45$, then we know both $x$ and $y$ are factors of 45, so $x = 3^{a_1} 5^{b_1}$ and $y = 3^{a_2} 5^{b_2}$ with $\max(a_1, a_2) = 2$ and $\max(b_1, b_2) = 1$.

We also know that $5x = 3y$, so $3^{a_1} 5^{b_1 + 1} = 3^{a_2 + 1} 5^{b_2}$. Which then implies that $a_1 > a_2$ and $b_2 > b_1$, forcing both of those to be equal to the maximums, i.e. $a_1 = 2$ and $b_2 = 1$ which is enough to give us the solution.

ConMan
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