Answer
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Hint: To solve this problem we first need to know that what are the composite numbers, composite numbers are the numbers which are completely divisible by at least one more number other than one and itself. So in the given question we have to explain why the given numbers are composite numbers so we should first find that if the given numbers have any other factors other than one and itself and if it does have then it is a composite number.
Complete step by step answer:
We are given the numbers 7 $\times $ 11 $\times $ 13 + 13 and 7 $\times $ 6 $\times $ 5 $\times $ 4 $\times $ 3 $\times $ 2 $\times $ 1 + 5 and we have to check whether they are composite numbers or not so,
First we should know that about composite numbers
Composite numbers are the numbers which are completely divisible by at least one other number other than one and itself.
So if we consider the first number 7 $\times $ 11 $\times $ 13 + 13 and simplify it to check if it has one more factor then it will be a composite number,
7 $\times $ 11 $\times $ 13 + 13
Taking 13 as common we get,
\[\begin{align}
& =13\left( 7\times 11+1 \right) \\
& =13\times \left( 77+1 \right) \\
& =13\times 78 \\
& =13\times 13\times 3\times 2 \\
& =1014 \\
\end{align}\]
So the given number is completely divisible by 13, 3 , 2, \[13\times 13=169\], etc. so the given number has more than two factor hence we proved that this number is a composite number.
Now the second number we have is 7 $\times $ 6 $\times $ 5 $\times $ 4 $\times $ 3 $\times $ 2 $\times $ 1 + 5, simplifying it as,
\[=7\times 6\times 5\times 4\times 3\times 2\times 1+5\]
Taking 5 as common we get,
\[\begin{align}
& =5\left( 7\times 6\times 4\times 3\times 2\times 1+1 \right) \\
& =5\left( 1008+1 \right) \\
& =5045 \\
\end{align}\]
So the given number has factors 5, 1009, etc. so the number of factors is greater than two it is also a composite number.
Note: You need to keep in mind that any two numbers which have more than two factors then it will be a composite number, else it will be a prime number. You can also directly check that if any number can be taken as common from the whole number then we can directly conclude it as a composite number for example in the above solution we have taken 13 and 5 common from the first and second numbers respectively so they are composite numbers.
Complete step by step answer:
We are given the numbers 7 $\times $ 11 $\times $ 13 + 13 and 7 $\times $ 6 $\times $ 5 $\times $ 4 $\times $ 3 $\times $ 2 $\times $ 1 + 5 and we have to check whether they are composite numbers or not so,
First we should know that about composite numbers
Composite numbers are the numbers which are completely divisible by at least one other number other than one and itself.
So if we consider the first number 7 $\times $ 11 $\times $ 13 + 13 and simplify it to check if it has one more factor then it will be a composite number,
7 $\times $ 11 $\times $ 13 + 13
Taking 13 as common we get,
\[\begin{align}
& =13\left( 7\times 11+1 \right) \\
& =13\times \left( 77+1 \right) \\
& =13\times 78 \\
& =13\times 13\times 3\times 2 \\
& =1014 \\
\end{align}\]
So the given number is completely divisible by 13, 3 , 2, \[13\times 13=169\], etc. so the given number has more than two factor hence we proved that this number is a composite number.
Now the second number we have is 7 $\times $ 6 $\times $ 5 $\times $ 4 $\times $ 3 $\times $ 2 $\times $ 1 + 5, simplifying it as,
\[=7\times 6\times 5\times 4\times 3\times 2\times 1+5\]
Taking 5 as common we get,
\[\begin{align}
& =5\left( 7\times 6\times 4\times 3\times 2\times 1+1 \right) \\
& =5\left( 1008+1 \right) \\
& =5045 \\
\end{align}\]
So the given number has factors 5, 1009, etc. so the number of factors is greater than two it is also a composite number.
Note: You need to keep in mind that any two numbers which have more than two factors then it will be a composite number, else it will be a prime number. You can also directly check that if any number can be taken as common from the whole number then we can directly conclude it as a composite number for example in the above solution we have taken 13 and 5 common from the first and second numbers respectively so they are composite numbers.
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