Question

For some natural number N, the number of positive integral x satisfying the equation, $1!+2!+3!+......+x!={{N}^{2}}$ is:
(a) None
(b) One
(c) Two
(d) Infinite

Answer 1

Hint: First, before proceeding for this, we must know the concept of the perfect square as the perfect square is the number is obtained by raising the power of the number by 2. Then, for this type of question, we need to use the hit and trial method to get our answer. Then, we use different values of x as 1, 2, 3, 4, 5 to get a condition and then conclude our answer.

Complete step-by-step solution:
In this question, we are supposed to find the number of times the value of the integral of the expression $1!+2!+3!+......+x!={{N}^{2}}$ gives a perfect square.
So, before proceeding for this, we must know the concept of the perfect square as the perfect square is the number is obtained by raising the power of the number by 2 as some of the examples are as:
$\begin{align}
  & {{1}^{2}}=1 \\
 & {{2}^{2}}=4 \\
 & {{3}^{2}}=9 \\
\end{align}$
Now, for this type of question, we need to use the hit and trial method to get our answer.
So, the first value of x, we suppose the first natural number as 1, we get:
$1!=1$
So, the value x=1 is one of the times the series $1!+2!+3!+......+x!={{N}^{2}}$ gives the perfect square of 1.
Then, we will go for the value of x as 2, we get:
$\begin{align}
  & 1!+2!=1+2\times 1 \\
 & \Rightarrow 1+2 \\
 & \Rightarrow 3 \\
\end{align}$
So, x=2 is not the natural number for the series $1!+2!+3!+......+x!={{N}^{2}}$ to get the perfect square.
Now, we try for the x=3, we get:
$\begin{align}
  & 1!+2!+3!=1+2\times 1+3\times 2\times 1 \\
 & \Rightarrow 1+2+6 \\
 & \Rightarrow 9 \\
\end{align}$
So, the value x=3 is the one of the times the series $1!+2!+3!+......+x!={{N}^{2}}$ gives the perfect square of 3.
Now, we try for the x=4, we get:
$\begin{align}
  & 1!+2!+3!+4!=1+2\times 1+3\times 2\times 1+4\times 3\times 2\times 1 \\
 & \Rightarrow 1+2+6+24 \\
 & \Rightarrow 33 \\
\end{align}$
So, x=4 is not the natural number for the series $1!+2!+3!+......+x!={{N}^{2}}$ to get the perfect square.
Now, we try for the x=5, we get:
$\begin{align}
  & 1!+2!+3!+4!+5!=1+2\times 1+3\times 2\times 1+4\times 3\times 2\times 1+5\times 4\times 3\times 2\times 1 \\
 & \Rightarrow 1+2+6+24+120 \\
 & \Rightarrow 153 \\
\end{align}$
Now, we try for the x=6, we get:
$\begin{align}
  & 1!+2!+3!+4!+5!+6!=1+2\times 1+3\times 2\times 1+4\times 3\times 2\times 1+5\times 4\times 3\times 2\times 1+6\times 5\times 4\times 3\times 2\times 1 \\
 & \Rightarrow 1+2+6+24+120+720 \\
 & \Rightarrow 873 \\
\end{align}$
So, after seeing the above pattern, we can conclude that for x>5 the series $1!+2!+3!+......+x!={{N}^{2}}$ gives the value of the form $10k+3$ where k is some natural number.
Also, we can say that the number whose unit digit is 3 can never be a perfect square in any case.
So, the natural number x equal to or greater than 5 doesn’t gives the perfect square.
So, only x=1 and x=3 satisfied the given equation as $1!+2!+3!+......+x!={{N}^{2}}$.
So, there are two cases for which the equation $1!+2!+3!+......+x!={{N}^{2}}$ satisfied for the perfect square condition.
Hence, option (c) is correct.

Note: Now, to solve these types of questions we need to know some of the basic things which are factorial in this question. So, to find the factorial of the number n, multiply the number n with (n-1) till it reaches 1. To understand let us find the factorial of 4.
$\begin{align}
  & 4!=4\times 3\times 2\times 1 \\
 & \Rightarrow 24 \\
\end{align}$