# Thread: Factoring numbers into product of primes

1. ## Factoring numbers into product of primes

Hey guys, need to confirm my answers for these following question.

Factoring the following numbers into the product of primes:

a) $\displaystyle 27$

= $\displaystyle 27 \cdot 1$

b) $\displaystyle 1024$

= $\displaystyle 2^{10}$

c) $\displaystyle 625000$

= $\displaystyle 2^3 \cdot 5^7$

d) $\displaystyle 12!$ (Also equals 479001600)

= $\displaystyle 2^{10} \cdot 3^5 \cdot 5^2 \cdot 7 \cdot 11$

Are my answers correct according to what the question is asking? thanks

2. Originally Posted by jvignacio
Hey guys, need to confirm my answers for these following question.

Factoring the following numbers into the product of primes:

a) $\displaystyle 27$

= $\displaystyle 27 \cdot 1$

b) $\displaystyle 1024$

= $\displaystyle 2^{10}$

c) $\displaystyle 625000$

= $\displaystyle 2^3 \cdot 5^7$

d) $\displaystyle 12!$ (Also equals 479001600)

= $\displaystyle 2^{10} \cdot 3^5 \cdot 5^2 \cdot 7 \cdot 11$

Are my answers correct according to what the question is asking? thanks
Sooooo close!

You got the harder ones right and the easy one wrong.

$\displaystyle 27 = 3^3$

Sooooo close!

You got the harder ones right and the easy one wrong.

$\displaystyle 27 = 3^3$
haha woops!!! Must of missed that one. Thank you for checking it!!