How do I answer a question when only one number is given and the other we need to find and the lcm and hcf are given

Question

How do I answer a question when only one number is given and the other we need to find and the lcm and gcd are given.

$X$ is a integer. The lcm of $x$ and $12$ is $120$. The $\gcd$ of $x$ and $12$ is $4$. Work the value of $x$.

Answer 1

1. You rely on the formula that $ab = \gcd(a,b)\cdot \operatorname{lcm}(a,b)$

So $12 x = \gcd(12,x)\operatorname{lcm}(12,x) = 120 \cdot 4$ for $x =\frac {120\cdot 4}{12} = 40$.

2. $\operatorname{lcm}(12,x) = 120=2^3\times 3\times 5$ so $x|120$ so the only possible prime divisors of $x$ are $2,3,5$ so $x = 2^a\times 3^b \times 5^c$. (So far as we know right now, $a,b,c$ might be $0$).

$12 = 2^2\times 3$.

We have $\gcd(12, x) = \gcd(2^2\times 3, 2^a\times 3^b \times 5^c) = 2^{\min(2,a)}\times 3^{\min(1,b)}\times 5^{\min(0,c)} = 4= 2^2$.

So $\min(2,a) =2$ so $a \ge 2$. $\min(1,b) = 0$ so $b = 0$. And $\min(0,c) = 0$ so.... we know nothing about $c$ yet.

We have $\operatorname{lcm}(12,x) =\operatorname{lcm}(2^2\times 3, 2^a\times 3^b \times 5^c)=2^{\max(2,a)}\times 3^{\max(1,b)}\times 5^{\max(0,c)}=120= 2^3\times 3^1\times 5^1$.

So $\max(2,a)=3$ so $a = 3$. $\max(1,b)=1$ so $b\le 1$ but we already knew $b-0$. $\max(0,c)=1$ so $c =1$.

So $x = 2^3\times 3^0\times 5^1=40$.