Question

See attached for the question about finding the prime factorization

Answer 1

a) We have a factorization but not all the numbers there are primes.

We can transform this as:

`\begin{gathered} 1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot9\cdot10\cdot11 \\ 1\cdot2\cdot3\cdot(2\cdot2)\cdot5\cdot(2\cdot3)\cdot7\cdot(2\cdot2\cdot2)\cdot(3\cdot3)\cdot(2\cdot5)\cdot11 \\ 1\cdot2^8\cdot3^4\cdot5^2\cdot7\cdot11 \end{gathered}`

Now we have the number expressed only with prime numbers.

b) We have to factorize the factors as:

`\begin{gathered} 6^4\cdot10\cdot49^(13) \\ (2\cdot3)^4\cdot(2\cdot5)\cdot(7^2)^(13) \\ 2^4\cdot3^4\cdot2\cdot5\cdot7^(26) \\ 2^5\cdot3^4\cdot5\cdot7^(26) \end{gathered}`

c) 191 is a prime number so it is already factorized.

d) We can factorize this number as:

`\begin{gathered} 10000^(12) \\ (10^4)^(12) \\ 10^(48) \\ (2\cdot5)^(48) \\ 2^(48)\cdot5^(48) \end{gathered}`