The product of four consecutive natural numbers is 5040. Find those numbers.
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
(x2+3x)(x2+3x+2)=5040
Let x2+3x=a
a(a+2)=5040
a2+2a=5040
Adding 1 on both sides,
a2+2a+1=5040+1
(a+1)2=5041
Taking square roots on both sides,
a+1=±71
x2+3x−70=0 or x2+3x+72=0
The discriminant of the 2nd equation is negative, hence no real roots exist for that equation. Hence, we only solve the 1st 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
∴x=−10 is not applicable
The numbers are 7,7+1=8,7+2=9,7+3=10
∴ The numbers are 7,8,9,10.