Question

Which of the following numbers is NOT prime?
A. $6!-1$
B. $6!-21$
C. $6!+41$
D. $7!-1$
E. $7!+11$

Answer 1

Hint: In this problem they have asked to find the numbers from the given options which are not prime. We can observe that the given options have the factorial values of integers. Now we will consider each option individually and evaluate the values of each option by using the formula $n!=n\left( n-1 \right)\left( n-2 \right)......3\times 2\times 1$. After that we will factorize the values of each option and check whether the obtained value is prime number or not.

Complete step-by-step solution:
Considering the first option which is $6!-1$.
We can write the value of $6!$ as
$\begin{align}
  & \Rightarrow 6!=6\times 5\times 4\times 3\times 2\times 1 \\
 & \Rightarrow 6!=720 \\
\end{align}$
Hence the value of $6!-1$ will be
$\begin{align}
  & \Rightarrow 6!-1=720-1 \\
 & \Rightarrow 6!-1=719 \\
\end{align}$
We can write the factors of $719$ as $1$, $719$. From the factors of $719$, we can say that the value $6!-1$ is a prime number.
Considering the second option which is $6!-21$.
We have calculated the value of $6!$ as $720$, hence the value of $6!-21$ will be
$\begin{align}
  & \Rightarrow 6!-21=720-21 \\
 & \Rightarrow 6!-21=699 \\
\end{align}$
We can write the factors of $699$ as $1$, $3$, $233$, $699$. From the factors of $699$, we can say that the value $6!-21$ is not a prime number.
Considering the third option which is $6!+41$.
We have calculated the value of $6!$ as $720$, hence the value of $6!+41$ will be
$\begin{align}
  & \Rightarrow 6!+41=720+41 \\
 & \Rightarrow 6!+41=761 \\
\end{align}$
We can write the factors of $761$ as $1$, $761$. From the factors of $761$, we can say that the value $6!+41$ is a prime number.
Considering the fourth option which is $7!-1$.
We can write the value of $7!$ as
$\begin{align}
  & \Rightarrow 7!=7\times 6\times 5\times 4\times 3\times 2\times 1 \\
 & \Rightarrow 7!=5040 \\
\end{align}$
Hence the value of $7!-1$ will be
$\begin{align}
  & \Rightarrow 7!-1=5040-1 \\
 & \Rightarrow 7!-1=5039 \\
\end{align}$
We can write the factors of $5039$ as $1$, $5039$. From the factors of $5039$, we can say that the value $7!-1$ is a prime number.
Considering the fifth option which is $7!+11$.
We have calculated the value of $7!$ as $5040$, hence the value of $7!+11$ will be
$\begin{align}
  & \Rightarrow 7!+11=5040+11 \\
 & \Rightarrow 7!+11=5051 \\
\end{align}$
We can write the factors of $5051$ as $1$, $5051$. From the factors of $5051$, we can say that the value $7!+11$ is a prime number.
Hence from all the values of options we can conclude that the value of the second option which is $6!-21$ not a prime number.

Note: For this problem we have got our result at the second option only. But in some cases, we don’t get the required result up to the last option. So, we need to check for each option even if we get the required result at the earliest options.