How to find gcf using prime factorization tree


Use Prime Factorization to Find GCF

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In my other lesson, I discussed the procedure on how to find the Greatest Common Factor using the List Method. This method is only effective when dealing with smaller numbers.

That’s why we need to learn a backup method to determine the GCF when larger numbers are involved. This alternate method takes advantage of the usefulness of Prime Factorization.

I must caution you that there is a prerequisite for this lesson. You will need to have an understanding of how to perform prime factorization on an integer.

Please don’t feel bad if you need to take the refresher lesson. Trust me, it is easy! You will soon realize that you are back here in no time. Here’s the link: Integer Prime Factorization


Steps on How to Determine the GCF using Prime Factorization

These are the steps on how to find the greatest common factor of two numbers using Prime Factorization. Although this method can be extended to find the GCF of multiple numbers, I just want to focus on two numbers. 2} and 5.

For the last step, we multiply together the numbers we have chosen from the previous step to determine the GCF. Thus, the GCF of 150 and 180 is 2 times 3 times 5 which is equal to 30.


Example 3: What is the GCF of 1,260 and 1,960?

Let’s take this skill of finding the greatest common factor to the next level. This time we will find the GCF of two numbers that have numerical values between 1,000 and 2,000.

By now, you’ve probably realized that finding the GCF of these numbers using the list method is going to be cumbersome. This is the reason why I have to create a separate lesson on finding the GCF using the Prime Factorization method. It is more efficient, accurate, and prone to less errors especially when you are working it out by hand.

To get to the heart of this method, you will need to have a good grasp on how to prime factorize a positive integer using the Prime Factor Tree.

I can’t overemphasize the importance of Prime Factor Tree. Trust me, it will be your best friend from here on out during your study of algebra in general.

In a nutshell, here’s the method of prime factorization using the Prime Factor Tree. Start dividing the given number by the smallest prime number which is 2. If 2 evenly divides the number, draw a diagonal down towards the left (branch of the tree) of the given number and write 2. Then, write the quotient by drawing a diagonal down towards the right. The quotient will become part of the trunk of the tree.

Keep dividing the subsequent quotients by 2 while recording your results as branches (divisors) and part of the trunk (quotients that are composite numbers). Keep going until such time when 2 can no longer divide the quotient. That’s when you move to the next prime number which is 3, and so on. Repeat the process until the quotient is a prime number. This is when you stop.

❖ Here’s the Prime Factor Tree of the number 1,260 and its Prime Factorization. 2} times 5 times 7 which gives us 140.


You might also be interested in:

Finding GCF using the List Method

Finding LCM using the List Method

Use Prime Factorization to Find LCM

Using Prime Factorizations - dummies

Every whole number greater than 1 has a prime factorization — that is, the list of prime numbers (including repeats) that equal that number when multiplied together. For example, here are the prime factorizations of 14, 20, and 300:

14 = 2 × 7

20 = 2 × 2 × 5

300 = 2 × 2 × 2 × 3 × 5

Factor trees are a useful tool for finding the prime factorization of a number. For example, to find the prime factorization 30, first find a factor pair — a pair of numbers that, when multiplied together, equals 30:

Note that 3 is circled because it's a prime number. But 10 is not prime, so you can continue the tree, finding a factor pair for 10:

This time, both 2 and 5 are circled because they're each prime numbers. When the bottom numbers on the factor tree are all prime, you have your answer: 30 = 2 × 3 × 5.

Using prime factorization to find the GCF

You can use prime factorization to find the greatest common factor (GCF) of a set of numbers. This method often works better for large numbers, when generating lists of all factors can be time-consuming.

Here's how to find the GCF of a set of numbers, using prime factorization:

  1. List the prime factors of each number.

  2. Circle every common prime factor — that is, every prime factor that's a factor of every number in the set.

  3. Multiply all the circled numbers.

    The result is the GCF.

For example, suppose you want to find the GCF of 28, 42, and 70. Step 1 says to list the prime factors of each number. Step 2 says to circle every prime factor that's common to all three numbers:

As you can see, the numbers 2 and 7 are common factors of all three numbers, so multiply these two numbers as follows:

2 × 7 = 14

Thus, the GCF of 28, 42, and 70 is 14.

Knowing how to find the GCF of a set of numbers is important when you begin reducing fractions to lowest terms.

Using prime factorization to find the LCM

One method for finding the least common multiple (LCM) of a set of numbers is to use the prime factorizations of those numbers. Here's how:

  1. List the prime factors of each number.

    Suppose you want to find the LCM of 18 and 24. List the prime factors of each number:

    18 = 2 × 3 × 3
    24 = 2 × 2 × 2 × 3
  2. For each prime number listed, underline the most repeated occurrence of this number in any prime factorization.

    The number 2 appears once in the prime factorization of 18 but three times in that of 24, so underline the three 2s:

    18 = 2 × 3 × 3
    24 = 2 × 2 × 2 × 3

    Similarly, the number 3 appears twice in the prime factorization of 18 but only once in that of 24, so underline the two 3s:

    18 = 2 × 3 × 3
    24 = 2 × 2 × 2 × 3
  3. Multiply all the underlined numbers.

    Here's the product:

    2 × 2 × 2 × 3 × 3 = 72

    So the LCM of 18 and 24 is 72. This solution checks out because

    18 × 4 = 72
    24 × 3 = 72

About This Article

This article is from the book:

  • Basic Math & Pre-Algebra For Dummies ,

About the book author:

Mark Zegarelli is a professional writer with degrees in both English and Math from Rutgers University. He has earned his living for many years writing vast quantities of logic puzzles, a hefty chunk of software documentation, and the occasional book or film review. Along the way, he’s also paid a few bills doing housecleaning, decorative painting, and (for ten hours) retail sales. He likes writing best, though.

This article can be found in the category:

  • Basic Math ,

How to factorize a number? – Wiki Reviews

So, what is the basic factorization of the number 5? What are the prime divisors of 5? We know that 5 is a prime number, which means that it has only two divisors, i.e. 1 and 5. So , the only prime divisor of 5 is the number itself.

What is the simple factorization of 9? The prime factorization of 9 is 9 = 3×3 .

Besides, what is a simple factorization of 8? The factors of the number 8 according to the simple factorization method are equal to 1, 2, 4 and 8 . Here 2 is a prime factor of 8.

Is a prime factorization of 32? Answer: The simple factorization of 32 is 2 × 2 × 2 × 2 × 2 = 2 5 .

What is prime factorization of 20?

Its main factors 1, 2, 4, 5, 10, 20 and (1, 20), (2, 10) and (4, 5) are pair factors.

What is the simple factorization of 3? 3 is the smallest odd prime number, and 3 is the only number equal to the sum of the preceding natural numbers. 3 is the only factor 3 together with 1. Both factors 3 is positive and is negative.
...
Factors 3.

1. What are factors 3?
4. Factors of 3 in pairs
5. The factors of the number 8 according to the simple factorization method are 1, 2, 4 and 8. Here 2 is a prime factor of the number 8.

What is the factorization of the number 4 into prime factors?

What is a prime factor of 4? Factors 4 1, 2 and 4 . 2 is the only prime factor of 4.

Also What is the prime factorization of 2? Factoring 2 using the simple factorization of

The simple factorization of 2 is 2 . 2 is the only even prime number. It has only two factors 1 and 2.

What is prime factorization of 31?

Coefficients 31 are 1 and 31 . It is only expressed as 31 × 1 or 1 × 31. Therefore, 31 is a prime number.

What is the simple factorization of 6? Solution: The factors of 6 by the simple factorization method are 1, 2, 3 and 6 . A prime number has no other divisors than 1 and itself. Therefore, the prime divisors of 6 are 2 and 3.

What is the factorization of 42 into prime factors?

Main factors 42 1, 2, 3 and 7 .

What is the factorization of the number 48 into prime factors?

Factors of 48 is a list of integers that can be divided by 48 without a remainder. There are 10 factors in total, of which 48 is the largest factor, and the prime factors of 48 are 2 and 3. 2 4 × 3 .

What is simple factorization 99? Solution. The prime divisors of 99 are 3 and 11. The product of all prime divisors of 99 is 33 .

What is the simple factorization of the numbers 6 and 12? To find GCF 6 and 12, we will find a simple factorization of given numbers, i.e. 6 = 2×3 ; 12 = 2x2x3.

What is factorization 4?

The prime factorization of 4 is 2 × 2 . A prime number has only two divisors: itself and 1.

What is the prime factorization of 18? So the simple factorization of 18 is 18 = 2 × 3 × 3 . The factor tree is not unique for a given number. Instead of expressing 18 as 2 × 9, we can express 18 as 3 × 6.

What is a factor tree?

Factor trees a way of expressing the divisors of a number , in particular, a simple factorization of a number. Each branch of the tree is divided into factors. As soon as the factor at the end of the branch is a prime number, there are only two factors left - itself and one, so the branch stops and we circle the number.

Is 1 a prime factor? The number 1 is called one. It has no prime factors and is neither prime nor compound.

What is prime factorization 33?

So 33 can be written as 33 = 3 × 11. So the prime factors of 33 are 3 and 11 .

What is the prime divisor of 6? The factors of 6 by the simple factorization method are 1, 2, 3, and 6. A prime number has no other divisor than 1 and itself. Therefore, the prime factors of 6 are 2 and 3 .

How to find a factor?

How to find the number of factors?

  1. Find its prime factorization, i.e. express it as a product of prime numbers.
  2. Write the prime factorization in exponent form.
  3. Add 1 to each of the exponents.
  4. Multiply all the numbers you get.
  5. This product will give the number of factors of the given number.

Factorization of Integers / Sudo Null IT News

Purpose of the article

The purpose of this article is to provide the reader with all the necessary information about the factorization of integers for its further use in bars with familiar programmers and mathematicians.

Why is this important?

Modern cryptography actively uses the RSA encryption algorithm. RSA itself is not practically reliable (semantically secured), since with the same values ​​of the input parameters (key and message) it produces the same result. However, it is convenient to use it as an auxiliary algorithm for encrypting, for example, a session key (used in TLS, as well as in early versions of PGP). It is easy to show that the problem of decrypting a message or a key encrypted with RSA is reduced to the problem of integer factorization.

Disclaimer

There is an excellent article by Connelly Barnes "Integer Factorization Algorithms". This article is not a direct translation of the one, with some additions and excluding code and evidence.

Materiel

A bit of information from an institute course in mathematics for a deeper understanding of the problem:

The fundamental theorem of arithmetic

Any natural number greater than one can be decomposed as a product of k prime numbers. For example . Now consider the number

The theorem is useful for limiting the number of iterations in factorization algorithms, as well as estimating their complexity. This theorem has a simple and beautiful proof:

If , then . Then , that is, and this is not possible.

An open problem in mathematics

At the moment, there is an open question in the theory of computational complexity: "Does the factorization of integers add to the complexity class P?". In simpler terms: Is it possible to find a divisor of the number N less than any given number in front of the given number in the number of steps , where is the number of digits , is the polynomial o. In other words, the answer to the question: "What is the fastest factorization algorithm?" not found yet.

Divisor enumeration

Divisor enumeration is the simplest factorization algorithm. Let's see if the number is natural for all s from 2 to .

The complexity of this algorithm - or depending on whether we run the algorithm over all numbers or only over primes.

Fermat's method

The method was proposed in 1600. It is based on a formula. Then instead of numbers we are looking for numbers . The numbers x and y are necessarily integers, since the sum and difference of odd numbers is an even number, and and (if p or s are even, then we can immediately solve the problem: s = 2).

In addition, and , and therefore . Then the algorithm looks like this: We successively check whether the number is the square of an integer for all x from to .

Worst case complexity of the algorithm . This algorithm has a number of improvements - optimization by divisor search, sieve method, Krajczyk-Fermat, etc. It is based on the following sequence:

It can be seen that the sequence is periodic and the period is, since everything is taken modulo N. It can be shown that if we find , then - the divisor of N. gcd is the greatest common divisor.

And this is true for any sequence with modulus N. In the article by Weisstein, Eric W. "Pollard Rho Factorization." it is shown that the expected period is proportional for almost all N. This is one of the reasons for using this one.

There are also other options for specifying the sequence: for example or .

Since we don't know s, Pollard suggested comparing and . Thus, the algorithm consists in sequentially checking all n for what is a non-trivial divisor of N. The complexity of such an algorithm is .

Brent's method

In 1980, Richard Brent published "An improved Monte Carlo factorization algorithm" in which he proposed an improvement to Pollard's ro-algorithm.

The difference is that instead we are looking for a divisor in the form , where m is the largest integer satisfying .

The sequence is also specified differently:

As in the Pollard algorithm, there are other different options for specifying such a sequence. The result of this improvement was a reduction in complexity: .

p-1 Pollard's factorization algorithm

In 1974, Pollard published another algorithm. It is based on Fermat's little theorem:

If p is a prime number that is a divisor of a natural number a, then .

To understand the algorithm, you need to take one more mathematical step: let there be a number such that - divisor. We can rewrite as by the factorial property. In this way: . We get that - divisor. We can look for a divisor in the form , and this is the algorithm.

In this case, it is difficult to explicitly define the complexity without going into complex mathematics, but conventionally the complexity can be written as .

Performance comparison of algorithms

The following graph shows the number of iterations of different algorithms for N different orders:

Performance of factorization algorithms

Output

The graph shows that factorization of order numbers requires about 1000 iterations. Finding a new, even more efficient algorithm can not only earn the creator $200,000, but also, perhaps, allow him to break modern encryption algorithms, and therefore improve them.


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