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

Go back to  'Ex.1.2'

## Question

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

Video Solution
Real Numbers
Ex 1.2 | Question 6

## 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.

Learn from the best math teachers and top your exams

• Live one on one classroom and doubt clearing
• Practice worksheets in and after class for conceptual clarity
• Personalized curriculum to keep up with school