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

$2^5" , "5^2$

$800 = 8xx100$

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

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

$2^3xx2^2xx5^2$

$2^5xx5^2$