Question

Find the prime factorization of 10!.

Short answer

$2^8·3^4·5^2·7$

Step-by-step solution

Step 1

By the definition of factorial:

$$\begin{gathered} 10!=10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1 \\ =10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2 \end{gathered}$$

2, 3, 5 and 7 are primes. Let us determine the prime factorization of all other factors.

To determine the prime factorization, we divide the given integer by successive primes. If the remainder is zero, then the prime is part of the factorization and we continue the prime factorization with the quotient.

$$\begin{gathered} 10=2\cdot 5 \\ 9=3\cdot 3 \\ 8=2\cdot 2\cdot 2 \\ 6=2\cdot 3 \\ 4=2\cdot 2 \end{gathered}$$

Step 2

We use these prime factorization in the above found product:

$$\begin{gathered} 10!=10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2 \\ =(2\cdot 5)\cdot (3\cdot 3)\cdot (2\cdot 2\cdot 2)\cdot 7\cdot (2\cdot 3)\cdot 5\cdot (2\cdot 2)\cdot 3\cdot 2 \end{gathered}$$

Finally combine like terms:

$10!=2^8·3^4·5^2·7$

Step 3

d) Factorize the given integer:

$101=1\times 101$

Since the factors are only 1 ad 101, 101 is prime.

e) Factorize the given integer:

$107=1\times 107$

Since the factors are only 1 and 107, 107 is prime.

f ) Factorize the given integer:

$113=1\times 113$

Since the factors are only 1 and 113, 113 is prime.