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