Question

210900 1,20018,000I need to decompose these numbers into the product of factors of prime numbers using the powers:

Answer 1

First factor the powers of 10 from the numbers. Since 10 can be decomposed as the product of prime numbers as 2*5, then, if a factor of 10^n is taken out from a number, it is the same as taking 2^*5^ out of the number.

Next, decompose the remaining factor as the product of prime numbers by repeated division by prime numbers.

210

First, factor out 10:

`210=21*10=21*2*5`

21 is not divisible by 2, but it is divisible by 3. Divide it to analyze the quotient:

`(21)/(3)=7`

7 is a prime number. Then, the prime fatorization of 210 is:

`210=2*3*5*7`

900

`900=9*100=9*10^2=9*2^2*5^2`

Since 9 is equl to 3^2, then, the prime factorization of 900 is:

`900=2^2*3^2*5^2`

1200

`1200=12*100=12*10^2=12*2^2*5^2`

12 is divisible by 2:

`(12)/(2)=6`

6 is divisible by 2:

`(6)/(2)=3`

3 is a prime number. Then, the prime factorization of 12 is 2*2*3. Then, the prime factorization of 1200 is:

`\begin{gathered} 1200=12*2^2*5^2 \\ =2*2*3*2^2*5^2 \\ =2^2*3*2^2*5^2 \\ =2^4*3*5^2 \end{gathered}`

18,000

`18,000=18*1000=18*10^3=18*2^3*5^3`

18 is divisible by 2:

`(18)/(2)=9`

9 is equal to 3^. The, the prime factorization of 18 is 2*3^2. Thn:

`\begin{gathered} 18,000=18*2^3*5^3 \\ =2*3^2*2^3*5^3 \\ =2^4*3^2*5^3 \end{gathered}`

Therefore, the prime factorization of the numbers 210, 900, 1200 and 18,000 are:

`\begin{gathered} 210=2*3*5*7 \\ 900=2^2*3^2*5^2 \\ 1200=2^4*3*5^2 \\ 18000=2^4*3^2*5^3 \end{gathered}`