Given: $\operatorname{gcf}(x,36)=9$ and $\operatorname{lcm}(x,36)=108$. Find $x$.

Question

Given: $\operatorname{gcf}(x,36)=9$ and $\operatorname{lcm}(x,36)=108$. Find $x$.

p.s I tried $54$ but although the $\operatorname{lcm}$ of $54$ and $36$ is $108$ the GCF is $18$ and it has to be $9$. Help me please.

Answer 1

Do you know $a\cdot b=$lcm$(a,b)\cdot$GCD$(a,b)$ where $a,b$ are positive integers (Demonstration )?

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Alternatively, as lcm $=10=2^2\cdot3^3$ so $x$ must be of the form $2^a3^b$ where $0\le a\le2,0\le b\le 3$

Now, as gcd$=9=3^2,$min$(a,2)=0\implies a=0,2$ being the highest power of $2$ in $36$

Similarly, $(2,b)=3\implies b=3$ which is the highest power of $3$ in $108$

Answer 2

Hint $\ $ It's immediate from the identity $\rm\, gcd(a,b)\, lcm(a,b) = ab.\,$ If this is not familiar then you can proceed as follows (essentially a special case of one way of proving said gcd * lcm identity).

$$\rm\ gcd(x,36)=9\!\!\overset{\large\rm\ \div\, 9}\iff gcd\left(\dfrac{x}9,4\right)=1\!\iff\! lcm\left(\dfrac{x}9,4\right) = 4\left(\dfrac{x}9\right)\!\!\!\overset{\large\,\ \times \,9}\iff\! \overbrace{lcm(x,36)}^{\large = \ \color{#c00}{108}} =\color{#c00}{4x}$$