Question

For some natural $N$ , the number of positive integral $x$ satisfying the equation,
$1!+2!+3!+......+\left(x\right)!=(N{)}^2$ is :

• A. None
• B. One
• C. Two
• D. Infinite

Answer 1

The correct option is C Two

$1!+2!+3!+......+\left(x\right)!=(N{)}^2$ is true for $x=1,x=3$ in that case $N=1,N=3$ respectively.

For, $x=4$, we have $1!+2!+3!+4!=33$ which is not a perfect square.

Again for $x\ge 5$ $1!+2!+....+\left(x\right)!$ is of the form $10k+3$ for $k$ is some natural number. In these cases, the given sum is not going to be a perfect square.

So, two values of $x$ satisfies the given equation.