SOLUTION: How do you find the product of primes for the number 10080?

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Question 793660: How do you find the product of primes for the number 10080?
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
To find the prime factorization for any number you look for prime factors.
First, is the number divisible by 2?
If so, divide it by 2, and keep on working with the result.
If not, try the same with 3; then with 5, 7, 11, 13. 17, 19, 23, and all the other prime numbers, until you get to one whose square exceeds the number.

10080 divides by 2 evenly and the result is 5040, so 10080=2%2A5040
5040 divides by 2 evenly and the result is 2520, so 10080=2%2A5040=2%2A2%2A2520
2520 divides by 2 evenly and the result is 1260, so 10080=2%2A5040=2%2A2%2A2520=2%2A2%2A2%2A1260
1260 divides by 2 evenly and the result is 630, so 10080=2%2A5040=2%2A2%2A2520=2%2A2%2A2%2A1260=2%2A2%2A2%2A2%2A630
We really do not have to keep writing the products, we just need to keep track of how many times we divide by 2.
630 divides by 2 evenly and the result is 315.
315 is not divisible by 2, but divides evenly by 3 and the result is 105.
105 divides evenly by 3 and the result is 35.
35 is not divisible by 3, but divides evenly by 5 and the result is 7.
So, all the factors that we peeled from 10080, like layers from an onion, are
10080=2%2A2%2A2%2A2%2A2%2A3%2A3%2A5%2A7=2%5E5%2A3%5E3%2A5%2A7
To check we can calculate back the product:
2%5E5=32 and 3%5E2=9, and 32%2A9%2A5%2A7=10080 according to my calculator.

Of course, if no one asks me to show my work, I take shortcuts, peeling larger numbers in easy chunks. For example I would divide by 10=2%2A5 in my head, to get 1008, and would look for factors of 1008, while keeping track of the 2 and 5 (I would write them on scrap paper).
I may realize that 1008 divides by 9=3%2A3, and dividing by 9 would get 112, keeping track of the two 3's in 9=3%2A3. Then I would work with 112 to find the remaining factors. Dividing by 4=2%2A2 I would get 28=4%2A7=2%2A2%2A7.
So the factor found were 2 and 5 first (as factors of 10),
3 and 3 next (in 9),
2 and 2 next (in 4),
and finally 2, 2, and 7 (in 28).
In all, 2 appeared 5 times, 3 appeared twice, 5 once, and 7 once,
so 10080=2%5E5%2A3%5E2%2A5%2A7