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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 $90.$

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

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  • $\begingroup$ 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. $\endgroup$ Commented Jun 27, 2022 at 20:32
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    $\begingroup$ I'm picking the factor with the smaller power each time. $2^1$ vs. $2^2$, $3^2$ vs $3^3$... Make sense? $\endgroup$
    – Doug
    Commented Jun 27, 2022 at 20:33
  • $\begingroup$ 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? $\endgroup$ Commented Jun 27, 2022 at 20:55
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    $\begingroup$ 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... $\endgroup$
    – Doug
    Commented Jun 27, 2022 at 20:58
  • $\begingroup$ Give yourself a change to practice: Find the HCF of 45 and 250. I will check your answer. $\endgroup$
    – Doug
    Commented Jun 27, 2022 at 21:00

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