# HCF and prime factorization [duplicate]

I'm following along in a math book I'm reading and the task at hand is to find the HCF of $$270$$ and $$900$$ using prime factorization. I know the answer is $$90$$ because I checked the answer at the back of the book and got it wrong.

I know that the only prime factors that go into each of them are $$2, 3$$ and $$5.$$ However I'm at a complete loss figuring out where to go from there to get $$90.$$

$$270 = 2\cdot 5\cdot3^3,\quad 900 = 2^2\cdot 5^2 \cdot 3^2.$$ The HCF is found by taking the smallest exponent of each distinct prime in the products. So, $$HCF(270,900) = 2\cdot 5\cdot 3^2 = 90.$$

• When you say exponent do you mean power? If yes, how do you determine which prime the exponent goes to and to what multiple, as in 3 to the power of two or three to the power of 3? I got it right at one stage but it was just from trial and error. Commented Jun 27, 2022 at 20:32
• I'm picking the factor with the smaller power each time. $2^1$ vs. $2^2$, $3^2$ vs $3^3$... Make sense?
– Doug
Commented Jun 27, 2022 at 20:33
• Sorry I'm not really following, let's say I have the list of common primes, in this instance 2, 3 and 5. What method do I need to follow to determine which way to multiply and exponentiate them in order to get the HCF? Commented Jun 27, 2022 at 20:55
• Let's say the number $2$ is common to both prime factorisations. Look at each number's prime factorisation. Choose the power of $2$ that is the larger of the two. This becomes one factor of the HCF. Proceed to the next distinct prime...
– Doug
Commented Jun 27, 2022 at 20:58
• Give yourself a change to practice: Find the HCF of 45 and 250. I will check your answer.
– Doug
Commented Jun 27, 2022 at 21:00