What is the prime factorization of 800?

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Answer 1 · 2016-12-03T19:25:36

#800 = 2xx2xx2xx2xx2xx5xx5= 2^5xx 5^2#

Start with any 2 factors of #800# and then break each down until you only have prime numbers as factors.

#800 = color(red)(8)xxcolor(blue)(100)#

#800 = color(red)(2xx4)xxcolor(blue)(10xx10)#

#800 = color(red)(2xx2xx2)xxcolor(blue)(2xx5xx2xx5)#

#800 =2xx2xx2xx2xx2xx5xx5#

You can also use "the ladder method"

Only divide by prime numbers until you get to #1#
You can divide in any order, but it is usually easier to divide any even number by 2 first

#2 |ul(800)#
#2 |ul(400)#
#2 |ul(200)#
#2 |ul(100)#
#2 |ul(50)#
#5 |ul(25)#
#5 |ul(5)#
#. |ul(1)#

#800 = 2xx2xx2xx2xx2xx5xx5= 2^5xx 5^2#

It does not matter which method you use, which factors you start with, or what order you divide in, the outcome will always be the same.

#20xx40 = 2^5 xx 5^2#

#16 xx50 = 2^5 xx 5^2#

Answer 2 · 2016-12-04T16:01:32

#2^5" , "5^2#

#800 = 8xx100#

#8" "=" "2xx4" " =" " 2xx2xx2" " =" " 2^3#

#100" "=" "10xx10" " =" " 2xx5xx2xx5" " =" " 2^2xx5^2#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Putting it all together")#

#2^3xx2^2xx5^2#

#2^5xx5^2#