# Ex.1.2 Q6 Real Numbers Solution - NCERT Maths Class 10

## Question

Explain why \(7 × 11 × 13 + 13\) and \(7 × 6 × 5 × 4 × 3 × 2 × 1 + 5\) are composite numbers.

## Text Solution

**What is unknown?**

Whether \(7{{ }} \times {{ }}11{{ }} \times {{ }}13{{ }} + {{ }}13\) and \(7{{ }} \times {{ }}6{{ }} \times {{ }}5{{ }} \times {{ }}4{{ }} \times {{ }}3{{ }} \times {{ }}2{{ }} \times {{ }}1{{ }} + {{ }}5\) are composite numbers?

**Reasoning:**

To solve this question, recall that:

- Prime numbers are whole numbers whose only factors are \(1\) and itself.
- Composite number are the positive integers which has factors other than \(1\) and itself.

Now, simplify \(7{{ }} \times {{ }}11{{ }} \times {{ }}13{{ }} + {{ }}13\) and \(7{{ }} \times {{ }}6{{ }} \times {{ }}5{{ }} \times {{ }}4{{ }} \times {{ }}3{{ }} \times {{ }}2{{ }} \times {{ }}1{{ }} + {{ }}5.\) On simplifying them, you will find that both the numbers have more than two factors. So, if the number has more than two factors, it will be composite.

**Steps:**

It can be observed that,

\[\begin{align} 7\times 11\times &13+13\,\\&=13\left( 7\times 11+1 \right) \\ \,\,&=13\left( 77+1 \right) \\ &=13\,\,\times \,78 \\ &=13\times 13\times 6\times 1 \\&=13\times13\times2\times3\times1\end{align}\]

The given number has \(2,3,13 \) and \(1\) as its factors. Therefore, it is a composite number.

\[\begin{align}7\times &6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5 \\&= {5 \times \left( {7\!\times\! 6 \!\times\! 4 \!\times\! 3 \!\times\! 2 \!\times\! 1 \!+\!1} \right)}\\&= {5 \times \left( {1008 + 1} \right)}\\ &={ 5 \times 1009 \times 1}\end{align}\]

\(1009\) cannot be factorised further. Therefore, the given expression has \(5,1009\) and \(1\) as its factors. Hence, it is a composite number.