Question

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

Answer 1

Let the four consecutive positive integers be x ,(x + 1), (x + 2) , (x + 3) 

∴ x (x+1) (x+2) (x +3) = 840 ...(given)

∴ x (x + 3) (x + 1) (x + 2) = 840

∴ (x² + 3x) [x (x+2)+1 (x+2)] = 840

∴ (x² + 3x) [x² + 2x + x +2] = 840 

∴ (x² + 3x) (x² + 3x +2) = 840

Let x² + 3x = m 

∴ m (m + 2) = 840

∴ m² + 2m = 840

∴ m² + 2m - 840 = 0    ....(+30 ,-28= -840)

∴ m² + 30m - 28m - 840 = 0

∴ m (m + 30) - 28(m + 30) = 0

∴ (m + 30) (m - 28) = 0

∴ m + 30 = 0        OR     m - 28 = 0

∴ m = -30                       m = 28

∵ We need positive integers.

∴ m ≠ -30        ∴ m = 28

Resubstituting m = x² + 3x 

∴ x² + 3x = 28 

∴ x² + 3x -28 = 0   

∴ x² + 7x - 4x - 28 = 0

∴ x (x + 7) - 4 (x +7) = 0

∴ (x + 7) (x - 4) = 0

∴ x + 7 = 0       OR    x - 4 = 0

∴ x = -7     OR    x = 4

∵ we need positive integers 

∴ x ≠ -7      but    x = 4

∴ The four consecutive positive integers are 4, 5, 6 and 7 respectively.