Question

How do you write 525 as a product of its prime factors?

Answer 1

`525=3xx5xx5xx7`

Explanation:

`2` is obviously not a factor of `525`
checking `3` we find:
`color(white)("XXX")525=3xx175`

`3` is not a factor of `175` so we continue on:
`4` is not a factor of `175` (since `2` wasn't)
checking `5` we find
`color(white)("XXX")175=5xx35`
`color(white)("XXX")rarr 525=3xx5xx35`

checking `5` again we find
`color(white)("XXX")35=5xx7`
`color(white)("XXX")rarr 525 = 3xx5xx5xx7`

Recognizing `7` as a prime, we can stop at this point.

Answer 2

`525 = 3*5*5*7`

Explanation:

Because we use base `10` to represent numbers, there are some simple rules to help identify whether numbers are divisible by `2`, `3` or `5`:

(1) A number is divisible by `2` when its last digit is even.

(2) A number is divisible by `3` when the sum of its digits is divisible by `3`.

(3) A number is divisible by `5` when its last digit is `0` or `5`.

In our example we find:

`525` is not divisible by `2` since its last digit `5` is odd.

`525` is divisible by `3` since `5+2+5 = 12` is divisible by `3`.

Dividing by `3` we find:

`525 = 3*175`

`175` is not divisible by `3` since `1+7+5 = 13` is not divisible by `3`.

`175` is divisible by `5` since its last digit is `5`.

Dividing by `5` we find:

`175 = 5 * 35`

`35` is divisible by `5` since its last digit is `5`.

Dividing by `5` we find:

`35 = 5*7`

`7` is prime so we can stop and review the result:

`525 = 3 * 175 = 3 * 5 * 35 = 3 * 5 * 5 * 7`