How to find all $m$ and $n$ so $gcd(m,n) = 36$ and $lcm(m,n) = 360 $?

Question

How to find all $m$ and $n$ so $gcd(m,n) = 36$ and $lcm(m,n) = 360 $?

Any suggestion would be helpful.

Answer 1

let $m=2^{p_1}3^{p_2}5^{p_3 }$ and $n=2^{q_1}3^{q_2}5^{q_3}$

but $\gcd(m,n)=36=2^23^2=2^{\,\min(p_1,q_1)}\;3^{\,\min(p_2,q_2)}\;5^{\,\min(p_3,q_3)} $

and $\operatorname{lcm}(m,n)=360=2^33^25^1=2^{\,\max(p_1,q_1)}\;3^{\,\max(p_2,q_2)}\;5^{\,\max(p_3,q_3)}$

Thus every $m,n$ such that

$\begin{cases} \min(p_1,q_1)=2 & \text{and} & \max(p_1,q_1)=3 \\ \min(p_2,q_2)=2 & \text{and} & \max(p_2,q_2)=2 \\ \min(p_3,q_3)=0 & \text{and} & \max(p_3,q_3)=1 \end{cases}$

satisfies the given relation.

Answer 2

Hint: Note that $gcd(n,m)\cdot lcm(n,m)=nm$. So we have $36\cdot 360=nm$.

Answer 3

Hint:

I suppose gcf means lcm for you.

If $a,b$ are two numbers and $d=\gcd(a,b)$, set $a'=\dfrac ad$, $b'=\dfrac bd$. Then $a',b'$ are coprime, and $$\operatorname{lcm}(a,b)=\frac{ab}{d}=a'b=ab'.$$

Therefore, you just have to find coprime $a',b'$ such that $$\frac{ab}{d}=a'b'd=360\iff a'b'=10. $$

Answer 4

We're looking for $m,n$ such that $c:= {\rm lcm}(m,n)=360$ and $d:=\gcd(m,n)=36$.

The nature of the least common multiple is that it is the multiple reduced by any common elements - that is, the greatest common divisor.

So you should know that $c=\frac{\large mn}{\large d}$, so $mn=cd = 12960$.

Also clearly $\frac{\large m}{\large d}\frac{\large n}{\large d}=\frac{\large c}{\large d}$. So taking $ a = \frac{\large m}{\large d}$ and $b = \frac{\large n}{\large d}$, we're looking for $ab = \frac{\large c}{\large d} = 10$.

This easily gives $(a,b) = \{(10,1),(5,2),(2,5),(1,10)\}$ and thus we have four options for $m,n$ (or two disregarding order): $(m,n) = \{(360,36),(180,72),(72,180),(36,360)\}$