Prime factorization

Revision as of 18:04, 11 June 2020 by Rockmanex3 (talk | contribs) (Updated link to Prime Shooter and added example)

For a positive integer $n$, the prime factorization of $n$ is an expression for $n$ as a product of powers of prime numbers. An important theorem of number theory called the Fundamental Theorem of Arithmetic tells us that every positive integer has a unique prime factorization, up to changing the order of the terms.

The form of a prime factorization is $n = {p_1}^{e_1} \cdot {p_2}^{e_2}\cdot{p_3}^{e_3}\cdots{p_k}^{e_k}$ where $n$ is any natural number, the $p_{i}$ are prime numbers, and the $e_i$ are their positive integral exponents.

Prime factorizations are important in many ways. One instance is to simplify fractions.


The common method of prime factorization is checking prime numbers, case by case. Use divisibility rules to check if primes (or powers of primes) are a factor and then move up to a different prime if said prime is not a factor. When a prime is a factor, we factor out the prime and check for factors in the resulting number.

Sometimes, we could easily see that a number is a multiple of a composite number or a prime. For instance, $50$ is a factor of $2650$, and $49$ is a factor of $4998$. In these cases, a good strategy is to choose the number accordingly to make factoring easier.


Consider the number $38052$.

  • Because the last two digits of the given number are a multiple of $4$, we can rewrite $38052$ as $2^2 \cdot 9513$. Note that $9513$ is not even, so we move on.
  • Because the sum of digits of $9513$ is a multiple of 9, we can rewrite $38052$ as $2^2 \cdot 3^2 \cdot 1057$. Note that the sum of digits of $1057$ is not a multiple of 3, so we move on.
  • Because $1057$ does not end in $0$ or $5$, we skip $5$ as a factor. As for $7$, after long division, we note that $1057 = 7 \cdot 151$, so we can rewrite $38052$ as $2^2 \cdot 3^2 \cdot 7 \cdot 151$. However, when doing long division again, $151$ is not a multiple of $7$, so we move on.
  • Note that $11$ is not a factor of $151$, so we can move on. In fact, because $13 > \sqrt{151}$, we can declare $151$ as prime and stop.

Thus, the prime factorization of $38052$ is $2^2 \cdot 3^2 \cdot 7 \cdot 151$.






See also

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