What are the factors of 128?

1 Answer
Mar 4, 2018

Prime factors: #128=2*2*2*2*2*2*2=2^7#

Regular factors: #1, 2, 4, 8, 16, 32, 64, 128#

Explanation:

We can use a factor tree and split up #128# until all factors we've found are prime:

#color(white)(..........................)128#
#color(white)(.........................) // color(white)(...)"\"#
#color(white)(........................) color(red)(2) color(white)(......)64#
#color(white)(..............................) // color(white)(.)"\"#
#color(white)(............................) color(red)(2) color(white)(....)32#
#color(white)(.................................) // color(white)(...)"\"#
#color(white)(...............................)color(red)(2)color(white)(....)16#
#color(white)(....................................) // color(white)(...)"\"#
#color(white)(..................................)color(red)(2)color(white)(.....)8#
#color(white)(........................................) // color(white)(.)"\"#
#color(white)(.......................................)color(red)(2)color(white)(.....)4#
#color(white)(.............................................) // color(white)(.)"\"#
#color(white)(...........................................)color(red)(2color(white)(....)2)#

Tallying up all the primes, we get:
#128=2*2*2*2*2*2*2=2^7#

If we want all the factors, not just the prime factors, we can obtain those by combining all the prime factors. In this case, all we have are two'2, so the combinations will just be all powers of two less than or equal to #7#:

#2^0, 2^1, 2^2, 2^3, 2^4, 2^5, 2^6, 2^7#

Computing all of the powers, we get:

#1, 2, 4, 8, 16, 32, 64, 128#