Find the prime factorisation of $6500$ and $1120$, and write down, in factorised form, $\gcd(6500, 1120)$ and $\operatorname{lcm}(6500, 1120)$. (i) Find the prime factorisation of $6500$, and of $1120$.
What is the typical way to go about this? Just using common divisibility rules? That's what I did. I'm not sure if there's a more structured way that I should be doing this, since it could be more difficult depending on the number? The above seem to be easy cases.
$65 \times 100 = 6500$
$13 \times 5 \times 25 \times 4 = 6500$
$13 \times 5 \times 5^2 \times 2^2 = 6500$
$13 \times 5^3 \times 2^2 = 6500$
$1120 = 112 \times 10$
$= 66 \times 2 \times 5 \times 2$
$= 6 \times 11 \times 2^2 \times 5$
$= 3 \times 11 \times 2^3 \times 5$
(ii) Hence write down, in factorised form, $gcd(6500, 1120)$ and $lcm(6500, 1120)$.
For GCD we just selected the highest powers of the numbers that are common to both? So it would be $\gcd(6500, 1120) = 5^3 \times 13 \times 2^3$?
And I think for LCM we take the lowest powers of each number? So it would be $\operatorname{lcm}(6500, 1120) = 5 \times 2^2$?
Thanks for any help.

One can do the gcd without factoring first. Oh, $$ \operatorname{lcm}(a,b) = \frac{ab}{\gcd(a,b)}  $$
$$ \frac{ 6500 }{ 1120 } = 5 +  \frac{ 900 }{ 1120 } $$ 
 $$ \frac{ 1120 }{ 900 } = 1 +  \frac{ 220 }{ 900 } $$ 
 $$ \frac{ 900 }{ 220 } = 4 +  \frac{ 20 }{ 220 } $$ 
 $$ \frac{ 220 }{ 20 } = 11 +  \frac{ 0 }{ 20 } $$ 
 Simple continued fraction tableau:
 $$ 
 \begin{array}{cccccccccc}
 & & 5 & & 1 & & 4 & & 11 & \\ 
  \frac{ 0 }{ 1 }   &   \frac{ 1 }{ 0 }   & &   \frac{ 5 }{ 1 }   & &   \frac{ 6 }{ 1 }   & &   \frac{ 29 }{ 5 }   & &   \frac{ 325 }{ 56 }  
 \end{array}
 $$ 
 $$  $$ 
 $$ 325 \cdot 5 - 56 \cdot 29 = 1 $$ 
$$  \gcd( 6500, 1120 ) = 20  $$
 $$ 6500 \cdot 5 - 1120 \cdot 29 = 20 $$ 
