Question

Find the prime factors using repeated division methods.
1. 22 2. 63 3. 35 4. 44 5. 49
6. 48 7. 50 8. 65 9. 87 10. 120

Answer 1

Hint: To find the prime factors, we will divide the number by the lowest prime number. We will repeat this until the number is no longer divisible by that number. Then we will divide it with the next prime number and repeat this process until the number is completely factored with all the factors as prime numbers.

Complete step-by-step answer:
First, let us consider 22.
We will find the factors of 22 with repeated division.
The lowers prime number is 2 and the number 22 is divisible by 22.
Therefore, 22 = 2 $\times $ 11
The number cannot be factorised further as 11 is a prime number.
Therefore, prime factors of 22 = 2 $\times $ 11 $\times $ 1
Next, we will factorise 63.
The lowest prime number that can divide 63 is 3.
Thus 63 = 3 $\times $ 21
On division, the quotient is 21, which is again divisible by 3.
So, 63 = 3 $\times $ 3 $\times $ 7
The quotient now is 7 and it is divisible by 7 itself, as 7 is a prime number.
Thus 63 = 3 $\times $ 3 $\times $ 7 $\times $ 1
Another way to represent continuous division is as follows:
$\begin{align}
  & 3\left| \!{\underline {\,
  63 \,}} \right. \\
 & 3\left| \!{\underline {\,
  21 \,}} \right. \\
 & 7\left| \!{\underline {\,
  7 \,}} \right. \\
 & \ \ 1 \\
\end{align}$
Thus, prime factors of 63 = 3 $\times $ 3 $\times $ 7 $\times $ 1
Now, we will see one more example.
Let us factorise 120.
The lowest prime number that can divide 120 is 2.
Thus, 120 = 2 $\times $ 60
The quotient is 60. It can also be divided by 2.
120 = 2 $\times $ 2 $\times $ 30
The quotient now is 30. It is also completely divisible by 2.
120 = 2 $\times $ 2 $\times $ 2 $\times $ 15
15 cannot be divided by 2, so we move to the next prime number, that is 3. It can completely divide 15.
120 = 2 $\times $ 2 $\times $ 2 $\times $ 3 $\times $ 5
The quotient 5 cannot be divided by 3, but it is divisible by 5 which is next prime number.
Thus, 120 = 2 $\times $ 2 $\times $ 2 $\times $ 3 $\times $ 5 $\times $ 1
In compact representation:
\[\begin{align}
  & 2\left| \!{\underline {\,
  120 \,}} \right. \\
 & 2\left| \!{\underline {\,
  60 \,}} \right. \\
 & 2\left| \!{\underline {\,
  30 \,}} \right. \\
 & 3\left| \!{\underline {\,
  15 \,}} \right. \\
 & 5\left| \!{\underline {\,
  5 \,}} \right. \\
 & \ \ 1 \\
\end{align}\]
Hence, prime factors of 120 = 2 $\times $ 2 $\times $ 2 $\times $ 3 $\times $ 5 $\times $ 1
Now, we will solve for other numbers in compact form.
3. 35
\[\begin{align}
  & 5\left| \!{\underline {\,
  35 \,}} \right. \\
 & 7\left| \!{\underline {\,
  7 \,}} \right. \\
 & \ \ 1 \\
\end{align}\]
Hence, prime factors of 35 = $5\times 7\times 1$.
4. 44
\[\begin{align}
  & 2\left| \!{\underline {\,
  44 \,}} \right. \\
 & 2\left| \!{\underline {\,
  22 \,}} \right. \\
 & 11\left| \!{\underline {\,
  11 \,}} \right. \\
 & \ \ \ \ 1 \\
\end{align}\]
Prime factors of 44 = $2\times 2\times 11\times 1$.
5. 49
\[\begin{align}
  & 7\left| \!{\underline {\,
  49 \,}} \right. \\
 & 7\left| \!{\underline {\,
  7 \,}} \right. \\
 & \ \ 1 \\
\end{align}\]
Prime factors of 49 = $7\times 7\times 1$.
6. 48
\[\begin{align}
  & 2\left| \!{\underline {\,
  48 \,}} \right. \\
 & 2\left| \!{\underline {\,
  24 \,}} \right. \\
 & 2\left| \!{\underline {\,
  12 \,}} \right. \\
 & 2\left| \!{\underline {\,
  6 \,}} \right. \\
 & 3\left| \!{\underline {\,
  3 \,}} \right. \\
 & \ \ 1 \\
\end{align}\]
Prime factors of 48 = $2\times 2\times 2\times 2\times 3\times 1$
7. 50
\[\begin{align}
  & 2\left| \!{\underline {\,
  50 \,}} \right. \\
 & 5\left| \!{\underline {\,
  25 \,}} \right. \\
 & 5\left| \!{\underline {\,
  5 \,}} \right. \\
 & \ \ 1 \\
\end{align}\]
Prime factors of 50 = $2\times 5\times 5\times 1$
8. 65
\[\begin{align}
  & 5\left| \!{\underline {\,
  65 \,}} \right. \\
 & 13\left| \!{\underline {\,
  13 \,}} \right. \\
 & \ \ \ \ 1 \\
\end{align}\]
Prime factors of 65 = $5\times 13\times 1$
9. 87
\[\begin{align}
  & 3\left| \!{\underline {\,
  87 \,}} \right. \\
 & 29\left| \!{\underline {\,
  29 \,}} \right. \\
 & \ \ \ \ 1 \\
\end{align}\]
Prime factors of 87 = $3\times 29\times 1$
10. 120
\[\begin{align}
  & 2\left| \!{\underline {\,
  120\,}} \right. \\
 & 2\left| \!{\underline {\,
  60 \,}} \right. \\
 & 2\left| \!{\underline {\,
  30 \,}} \right. \\
 & 3\left| \!{\underline {\,
  15 \,}} \right. \\
 & 5\left| \!{\underline {\,
  5 \,}} \right. \\
 & \ \ 1 \\
\end{align}\]
Prime factors of 12 = $2\times 2\times 3\times 5$

Note: Any numbers can be factored in the similar manner, either using compact representation or textual representation. A number will always be divisible by any of its prime factors and also a combination of factors.