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

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Question

Explain why \(\begin{align}7 × 11 × 13 + 13\end{align}\) and \(\begin{align}7 × 6 × 5 × 4 × 3 × 2 × 1 + 5\end{align}\) 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.

  
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