# Factoring numbers into product of primes

• Apr 13th 2010, 03:44 AM
jvignacio
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
• Apr 13th 2010, 04:48 AM
Quote:

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.

$\displaystyle 27 = 3^3$
• Apr 13th 2010, 05:17 AM
jvignacio
Quote:

$\displaystyle 27 = 3^3$