Write Numbers as Products of Prime FactorsDate: 10/23/96 at 17:40:57 From: Antonio J. Filipe Subject: Write as products of prime factors I need to write 4, 6, 8, 10, 12, and 14 as products of their prime factors. Please help! My mind is a blank. Date: 01/04/97 at 15:46:18 From: Doctor Jaime Subject: Re: Write as products of prime factors Recall that prime numbers are divisible only by themselves and one. Examples are: 2, 3, 5, 7, 11, 13, etc. The prime factors you need are prime numbers such that their product is the given number. So you need to do two things: 1. find the prime numbers that are factors of the given number; 2. write the given number as a product of the prime factors found. To obtain the prime factors you just have to divide the given number by the prime numbers and see which divide the given number (with zero remainder). You begin by using the smaller prime numbers, because they are easier, to see which ones divide your number. When you find a prime number that divides your number, you get a certain quotient and you then begin the same process with the quotient. For example, let's try this with number 12. We begin with the smallest prime number: 2. 12 divided by 2 gives 6 as a quotient and zero as a remainder. So 2 divides 12, and we can say that 12 = 2 x 6. We have now written 12 as a product of two numbers, 2 and 6, but they are not both prime factors. 2 is a prime number, but 6 is not. Therefore, we now need to write 6 as a product of prime factors. Again we begin with the smallest prime number: 2. 6 divided by 2 gives 3 as a quotient and zero as a remainder. So 2 divides 6, and we can say that 6 = 2 x 3. The divisions end here because the quotient obtained, 3, is a prime number (it can only be divided by itself and give quotient 1). Now, how do we write 12 as a product of its prime factors (2 and 3 in this case)? Above we have obtained 12 = 2 x 6 = 2(2 x 3) = 2^2 x 3, (where 2^2 means 2 squared or 2 raised to the second power). So it must be: 12 = 2^2 x 3 Let's try again with number 45, for example. We begin with the smallest prime number: 2. 45 divided by 2 does not give zero as a remainder, so we must move to the next prime number, 3. 45 divided by 3 gives 15 as a quotient and zero as a remainder, so 3 divides 45, and we can say that 45 = 3 x 15. 15 divided by 3 gives 5 as a quotient and zero as a remainder. So 3 divides 15, and we can say that 15 = 3 x 5. The divisions end here because the quotient obtained, 5, is a prime number. So 3 and 5 are the only prime factors of 45. Now, how do we write 45 as a product of its prime factors (3 and 5 in this case)? Above we have obtained 45 = 3 x 15 = 3 * (3*5) = 3^2 * 5. So it must be: 45 = 3^2 * 5 You can again verify that it is really so, doing the calculation. Another method to find the prime factors is to use the so-called "factor trees." You find factors of your number, and then factors of these factors, etc., until you arrive at prime numbers. For example, if you want to factor 24, you may know that 24 = 6 x 4. But 6 = 2 x 3 and 4 = 2 x 2, and all the factors are prime. You can write this as a "tree": 24 / \ 6 4 / \ / \ 2 3 2 2 So the prime factors are 2 and 3. But 2 shows up three times and 3 shows up once, so: 24 = 2^3 x 3 For another answer with some more information, see http://mathforum.org/dr.math/problems/primefac.html -Doctor Jaime, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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