Question

product of 4 consecutive natural number is 5040. Find the numbers?
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Answer 1

Answer:

numbers are 7,8,9,10.

Step-by-step explanation:

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 = 5040                                              [ a^2 + 2ab + b^2 = (a+b)^2 ]

Taking square roots on both sides,

a+1=±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  

∴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.