Question

The product of four consecutive positive integers is 840 . Find the largest number .

Answer 1

Let the four consecutive positive numbers be x, (x + 1), (x + 2) and (x + 3).
Then, by the given condition, we get:
x (x + 1)(x + 2)(x + 3) = 840
$\Rightarrow$ {x(x + 3)}{(x + 1)(x + 2)} = 840
$\Rightarrow$ (x² + 3x)(x² + 3x + 2) = 840
Let x² + 3x = m. Then, we get:
m(m + 2) = 840
$\Rightarrow$ m² + 2m – 840 = 0
On splitting the middle term 2m as 30m – 28m, we get:
m² + 30m – 28m – 840 = 0
$\Rightarrow$ m(m + 30) – 28(m + 30) = 0
$\Rightarrow$ (m + 30)(m – 28) = 0
$\Rightarrow$m + 30 = 0 or m – 28 = 0
$\Rightarrow$ m = –30 or m = 28
On substituting m = x² + 3x, we get:
$\Rightarrow$ x² + 3x = –30 or x² + 3x = 28
$\Rightarrow$x² + 3x + 30 = 0 …(1) or x² + 3x – 28 = 0 …(2)

In equation (1), we have a = 1, b = 3, c = 30.
We know that the discriminant of a quadratic equation is Δ = b² – 4ac.
Thus, we get:
Δ = (3)² – 4(1)(30)
$\Rightarrow$Δ = 9 – 120 = –111
Since Δ < 0, the roots of the quadratic equation (1) are not real.

In equation (2), we have a = 1, b = 3, c = –28.
We know that the discriminant of a quadratic equation is Δ = b² – 4ac.
Thus, we get:
Δ = (3)² – 4(1)(–28)
$\Rightarrow$ Δ = 9 + 112 = 121
Since Δ > 0, the roots of the quadratic equation (2) are real.

Now, from equation (2), we get:
x² + 3x – 28 = 0
$\Rightarrow$x² + 7x – 4x – 28 = 0
$\Rightarrow$ x(x + 7) – 4(x + 7) = 0
$\Rightarrow$ (x + 7)(x – 4) = 0
$\Rightarrow$ x + 7 = 0 or x – 4 = 0
$\Rightarrow$x = –7 or x = 4
Since x is a positive number, x = 4.
Thus, the largest number is x + 3 = 4 + 3 = 7