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:

$210=21\times10=21\times2\times5$

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

$\frac{21}{3}=7$

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

$210=2\times3\times5\times7$

900

$900=9\times100=9\times10^2=9\times2^2\times5^2$

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

$900=2^2\times3^2\times5^2$

1200

$1200=12\times100=12\times10^2=12\times2^2\times5^2$

12 is divisible by 2:

$\frac{12}{2}=6$

6 is divisible by 2:

$\frac{6}{2}=3$

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

$\begin{gathered} 1200=12\times2^2\times5^2 \\ =2\times2\times3\times2^2\times5^2 \\ =2^2\times3\times2^2\times5^2 \\ =2^4\times3\times5^2 \end{gathered}$

18,000

$18,000=18\times1000=18\times10^3=18\times2^3\times5^3$

18 is divisible by 2:

$\frac{18}{2}=9$

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

$\begin{gathered} 18,000=18\times2^3\times5^3 \\ =2\times3^2\times2^3\times5^3 \\ =2^4\times3^2\times5^3 \end{gathered}$

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

$\begin{gathered} 210=2\times3\times5\times7 \\ 900=2^2\times3^2\times5^2 \\ 1200=2^4\times3\times5^2 \\ 18000=2^4\times3^2\times5^3 \end{gathered}$