What is the prime factorization of 491?

1 Answer
Nov 30, 2016

It's prime factorization is itself, #491=491xx1#

Explanation:

491 is a prime number since it's not divisible by 2, 3, 5, 7, 11, 13.. or any other prime number that is smaller than 491.

Instead of testing all the prime numbers up 491 there's a quick way of doing it.

The concept is simple, if a number N can be written as #N=axxb# there are 3 cases, it's either :

  • #a=sqrt(N) and b=sqrt(N)#
  • #a < sqrt(N) and b > sqrt(N)#
  • #a > sqrt(N) and b < sqrt(N)#

#a and b# CAN'T be both bigger than #sqrt(N)# because then we'll have #axxb>N#

So we'll only need to test prime numbers smaller than #sqrt(N)#, if N is divided by one of them then it's not prime.

#sqrt(491)=22,15# so we only need to test the prime numbers :

2, 3, 5, 7, 11, 13, 17, and 19

Using a calculator you'll find that 491 has no factors among theses numbers so it's a prime number itself.

Sources :
(http://stackoverflow.com/questions/5811151/why-do-we-check-up-to-the-square-root-of-a-prime-number-to-determine-if-it-is-pr)
(http://www.counton.org/explorer/primes/checking-if-a-number-is-prime/)