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Prime Numbers

What is meant by Prime Numbers? 

The numbers that have only two factors i.e. \(\textstyle 1 \) and \(\text{the number itself}\) are known as Prime Numbers. So, there are \(25\) prime numbers between \(1\) and \(100\), i.e. 

\(2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97\)

Let’s understand this with the help of some examples:

Numbers   Factors
\(1 \) \(1 × 1\) \(1 \)
\(2 \) \(1 × 2\) \(1, 2\)
\(3 \) \(1 × 3\) \(1, 3\)
\(4 \) \(1 × 4\\ 2 × 2\) \(1, 2, 4\)

Look at the table above. Which numbers have exactly \(2\) factors?

The answer would be \(2\) and \(3\).

So, a prime number is a whole number that has exactly \(2\) factors, \(1\) and itself.

Example: \(2, 3, 5, 7, 11, 13, 17\) and so on.

 

Composite Numbers

A composite number is a whole number that has more than two factors.

Example:

\(4, 6, 8, 9, 12, 14, 15\) and so on.

Numbers   Factors
\(5 \) \(1 × 5\) \(1, 5\)
\(6 \) \(1 \times 6\\ 2 \times 3\) \(1, 2, 3, 6\)
\(7 \) \(1 × 7\) \(1, 7\)
\(8 \) \(1 × 8\\ 2 × 4\) \(1, 2, 4, 8\)

Look at the table above. Which numbers have more than \(2\) factors?

The answer would be \(6\) and \(8\).

 

Prime Factorisation

Example:

Express \(12\) as a product of its prime factors.

Number   Prime Factor   Prime Factor   Prime Factor
\(12 \) \(= \) \(2 \) \(\times\) \(6 \)    
  \(= \) \(2 \) \(\times\) \(2 \) \(\times\) \(3 \)

\(12 = 2 × 2 × 3\) or \(12 = 2^2 × 3\)

 

Question 1:

Express \(175\) as a product of its prime factors, leaving your answers in index notation.

Solution: 

\(\begin{array}{c|lcr} 5 & 175 \qquad \\ \hline 5 & 35 \\ \hline 7 & 7 \\ \hline & 1 \end{array} \)

\(\begin{align*} 175 &= 5 × 5 × 7\\ &= 5^2 × 7 \end{align*}\)

 

Square Roots And Cube Roots

To find the square root of a number, divide the index of each prime factor by 2.

Let’s understand this with the help of some examples:

Using a calculator, \(\sqrt{16} = 4\); why? 

Method 1:

\( \begin{align*} 4 × 4 = 16\\ 4^2 = 16 \end{align*}\)
    \(\begin{align*} 16 &= 4^2\\      &= 4 \end{align*}\)

 Method 2:

\(\begin{align*} 2 × 2 × 2 × 2 = 16\\ 2^4 = 16 \end{align*}\)
              \(\begin{align*} \sqrt{16} &= 24\\       &= 2^2\\     &= 4 \end{align*}\)

 

To find the cube root of a number, divide the index of each prime factor by \(3\)

Let’s understand this with the help of some examples:

 

Question 2:

Using calculator, \(\sqrt[3]{64} = 4 \). Why? 

Solution: 

Method 1:

\(\begin{align*} 4 × 4 × 4 = 64\\ 4^3 = 64 \end{align*}\)
            \(\begin{align*} \sqrt[3]{64} &= \sqrt[3]{4^3}\\ &= 4 \end{align*}\)

Method 2: 

\(\begin{align*} 2 × 2 × 2 × 2 × 2 × 2 = 64\\ 2^6 = 64 \end{align*} \)
 \(\begin{align*} \sqrt[3]{64} &= \sqrt[3]{2^6}\\ &=2^2\\ &= 4 \end{align*}\)

 

Highest Common Factor (HCF)

Number Factors
\(12\) \(1, 2, 3, 4, 6, 12\)
\(18 \) \(1, 2, 3, 6, 9, 18\)

The highest common factor of \(12\) and \(18\) is \(6\).

To find the HCF of two or more numbers, multiply the common prime factors with the lowest index together.

 

Question 3:

Find the highest common factor (HCF) of \(55\)\(165\) and \(605\).

Solution: 

Step 1: Prime factorization

\(\begin{align*} 55 &= 5 × 11\\ 165 &= 3 × 5 × 11\\ 605 &= 5 × 11^3 \end{align*}\)

 

Step 2: Identify common prime factors

\(5 \;\text{and} \;11\)

 

Step 3: Multiply the common prime factors with the lowest index.

\(\begin{align*} HCF &= 5 × 11\\          &= 55 \end{align*}\)

 

Lowest Common Multiple (LCM)

Number Multiples
\(3\) \(3, 6, 9, 12, 15, 18\)
\(4 \) \(4, 8, 12, 16, 20\)

The lowest common multiple of \(3\) and \(4\) is \(12\).

To find the LCM of two or more numbers, multiply the unique prime factors with the highest index together.

 

Question 4:

Find the lowest common multiple (LCM) of \(18, 63 \;and \;81\).

 

Solution: 

Step 1: Prime factorization

\(\begin{align*} 18 &= 2 × 3^2\\ 63 &= 3^2 × 7\\ 81 &= 3^4 \end{align*}\)

 

Step 2: Identify the unique prime factors

\(2, 3, and \;7\)

 

Step 3: Find the highest index of each prime factor

\(\begin{align} \text{LCM} &= 2 × 3^4 × 7\\ &= 1134 \end{align}\)

Continue Learning
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Basic Algebra And Algebraic Manipulation 1 Approximation And Estimation
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