I'm getting through a HCF problem and I'm having trouble taking the primes and determining what way to string them to sum up to the HCF. Here's what I have
270 and 900 are the targets
270 = 2 X 3 X 3 X 3 X 5
900 = 2 X 2 X 3 X 3 X 5 X 5
The HCF is 90, I came to that conclusion by repeatedly guessing combinations of the prime factors, I got the answer but there was no logic or process to it. Where do I go from the stage I'm at to the answer?
$$270 = 2^\color{blue}{1}\times 3^3\times 5^{\color{blue}{1}}$$ $$900 = 2^2\times 3^{\color{blue}{2}}\times 5^2$$
You ask how to find the HCF (also notated as the gcd) of $270$ and $900$? You can do this by inspection by looking at the prime factorizations and taking as many copies of each prime as the number who had the fewer of that prime present. Here, $2$ appears in $270$ once and twice in $900$ so we will use just one copy of $2$. $3$ appears three times in $270$ and twice in $900$ so we'll use two copies of $3$. Similarly $5$ once in $270$ and twice in $900$ so we'll use one copy of $5$. These are the only primes in use, giving a final answer of $2\times 3\times 3\times 5 = 90$.
$$90 = 2^\color{blue}{1}\times 3^\color{blue}{2}\times 5^\color{blue}{1}$$
Recall that a prime missing from one of the numbers is equivalent to the prime appearing with its power equal to zero
A more formal method is the Euclidean Division Algorithm.
Starting with $900$ and $270$ we ask what the largest integer multiple is of $270$ (the smaller) which is smaller than $900$ (the larger). That would be $3\times 270 = 810$. Since we know (with $a\neq b$) that $\gcd(a,b)=\gcd(a-b,b)$ this implies that $\gcd(900,270)=\gcd(900-3\cdot 270,270)=\gcd(90,270)$. From there it is clear that $90$ is a divisor of $270$ and so the HCF is $90$ itself.
