Question

Explain why $ 7\times 11\times 13+13 $ and $ 7\times 6\times 5\times 4\times 3\times 2\times 1+5 $ are composite numbers. \[\]

Answer 1

Hint: We recall the definitions of composite number, prime number and the distributive property of multiplication and division as $ a\times \left( b+c \right)=a\times b+a\times c $ to take 13 common from $ 7\times 11\times 13+13 $ and 5 common from $ 7\times 6\times 5\times 4\times 3\times 2\times 1+5 $ . Since 56 and 13 become one more factor of the numbers $ 7\times 11\times 13+13 $ and $ 7\times 6\times 5\times 4\times 3\times 2\times 1+5 $ respectively and then those numbers are composite. \[\]

Complete step by step answer:
We know that when a dividend is exactly dividend the divisor we call that divisor is a factor of the dividend. We know that we call number prime when it has two factors 1 and the number itself. We call a number composite when it has more than two factors.\[\]
We know from the distributive property of multiplication and division that $ a\times b+a\times c=a\times \left( b+c \right) $ . The process of taking $ a $ outside the bracket is called the process of taking $ a $ common. \[\]
We are given two numbers $ 7\times 11\times 13+13 $ and $ 7\times 6\times 5\times 4\times 3\times 2\times 1+5 $ and are asked to explain why they are composite. Let us consider the first number . We have;
\[\begin{align}
  & \Rightarrow 7\times 11\times 13+13 \\
 & \Rightarrow 7\times 11\times 13+13\times 1 \\
\end{align}\]
We take 13 common in the above step to have;
\[\begin{align}
  & \Rightarrow 13\left( 7\times 11+1 \right) \\
 & \Rightarrow 13\times \left( 77+1 \right) \\
 & \Rightarrow 13\times 78 \\
\end{align}\]
We see that except one and the number itself $ 7\times 11\times 13+13 $ has two more factors 13 and 78. Since now it has more than two factors $ 7\times 11\times 13+13 $ is a composite number. Let us consider the second number. We have;
\[\begin{align}
  & \Rightarrow 7\times 6\times 5\times 4\times 3\times 2\times 1+5 \\
 & \Rightarrow 7\times 6\times 5\times 4\times 3\times 2\times 1+5\times 1 \\
\end{align}\]
We take 5 common in the above step to have;
\[\begin{align}
  & \Rightarrow 7\times 6\times 5\times 4\times 3\times 2\times 1+5 \\
 & \Rightarrow 7\times 6\times 5\times 4\times 3\times 2\times 1+5\times 1 \\
 & \Rightarrow 5\left( 7\times 6\times 4\times 3\times 2\times 1+1 \right) \\
 & =5\times \left( 1008+1 \right)=5\times 1009 \\
\end{align}\]
 We see that except one and the number itself $ 7\times 6\times 5\times 4\times 3\times 2\times 1+5 $ has two more factors 5 and 1009. Since now it has more than two factors $ 7\times 6\times 5\times 4\times 3\times 2\times 1+5 $ is a composite number.\[\]

Note:
 We can find all the factors of the number of a number by taking product combinations from the prime factorization of that number. The prime factors of $ 7\times 11\times 13+13 $ are $ 1,2,3,13 $ and prime factors of $ 7\times 6\times 5\times 4\times 3\times 2\times 1+5 $ are 5 and 1009. A composite number except 1 and the number itself may have prime or composite factors. We should always remember divisibility rules to check whether a prime or composite can exactly divide a number.