Question

The product of four consecutive natural numbers is $5040$. Find those numbers.

Answer 1

Let the four consecutive natural numbers be $x,x+1,x+2,x+3$.
According to given condition,
$x(x+1)(x+2)(x+3)=5040$
$x(x+3)(x+1)(x+2)=5040$
$(x^2+3x)(x^2+3x+2)=5040$
Let $x^2+3x=a$
$a(a+2)=5040$
$a^2+2a=5040$
Adding $1$ on both sides,
$a^2+2a+1=5040+1$
$(a+1{)}^2=5041$
Taking square roots on both sides,
$a+1=\pm 71$
$x^2+3x-70=0$ or $x^2+3x+72=0$

The discriminant of the $2^{nd}$ equation is negative, hence no real roots exist for that equation. Hence, we only solve the $1^{st}$ equation.
$x(x+10)-7(x+10)=0$
$(x+10)(x-7)=0$
$x+10=0$ or $x-7=0$
$x=-10$ or $x=7$
$-10$ is not a natural number
$\therefore x=-10$ is not applicable
The numbers are $7,7+1=8,7+2=9,7+3=10$
$\therefore$ The numbers are $7,8,9,10$.