Question

The product of three consecutive numbers when divided by each of them in turn, the sum of the
three quotients will be 74. What are the numbers?
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Answer 1

Solution

Let the numbers be x, x + 1 and x + 2.

Product of consecutive number = x(x + 1)(x + 2)

$\tiny{\frac{x(x + 1)(x + 2)}{x} + \frac{x(x + 1)(x + 2)}{x + 1} + \frac{x(x + 1)(x + 2)}{x + 2} = 74}$

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➵(x+1)(x+2) + x(x+2) + x(x+1) = 74

➵x² + 3x + 2 + x² + 2x + x² + x= 74

➵ 3x² + 6x + 2 - 74 = 0

➵ 3x² + 6x - 72 = 0

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D = b² - 4ac

D = 36 - 4(3)(-72)

D = 36 + 864

D = 900

Shridharacharya Method

➵ x = (- b ± √D)/2a

➵ x = (- 6 ± √900)/6

➵ x = [- 6 ± (30)]/6

Case 1 : If it is negative,

➵ x = (- 6 - 30)/6

➵ x = - 6

Case 1 : If it is positive,

➵ x = (- 6 + 30)/6

➵ x = 4

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So if x is - 6,

Then, x + 1 = - 5

and x + 2 = - 4

&

Or if x is 4,

Then, x + 1 = 5

and x + 2 = 6

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Numbers :

- 6, - 5 and - 4 Or 4, 5 and 6

Answer 2

Solution :-

$\textsf{Let the three number of be x, x - 1 and x + 1.}$

ATQ,

$\sf x(x+1) + (x-1)(x+1) + x(x-1) = 74 \\ \sf</p><p> \Rightarrow x^2+x + x^2-1 + x^2-x=74 \\ \sf </p><p>\Rightarrow 3x^2 = 75 \\ \sf</p><p>\Rightarrow x^2 = 25 \\ \sf</p><p>\Rightarrow x = 5 \: or \: -5$

$\textsf{Three consecutive numbers are (-6, -5, -4 ) or ( 4, 5, 6 )}$