Question

. Three consecutive integers are such that their product is three times the middle number
multiplied by 40. Find the absolute value of the largest number.​

Answer 1

Answer:

(x)+(x+1)+(x+2)=(x+1)40

3x+3=40x+40

40x-3x=40-3

37x=37

x=37/37

x=1

Answer 2

$Let \: (x-1) , x \:and \:(x+1) \:are \: three \\consecutive \: integers$

/* According to the problem given */

$(x-1)x(x+1) = 40 \times 3 x$

$\implies (x-1)(x+1)x = 120x$

$\implies (x-1)(x+1)x - 120x = 0$

$\implies x[ (x+1)(x-1) - 120 ] = 0$

$\implies x( x^{2} - 1 - 120 ) = 0$

$\implies x( x^{2} - 121 ) = 0$

$\implies x = 0 \:Or \: x^{2} - 121 = 0$

$\implies x = 0 \:Or \: x^{2} = 121$

$\implies x = 0 \:Or \: x^{2} = 11^{2}$

$\implies x = 0 \:Or \: x^{2} = \pm 11$

$Absolute \: value \: largest \: number \\= ( x+1 ) \\= 11 + 1 \\= 12$

Therefore.,

$\red{Absolute \: value \: largest \: number}\green {= 12 }$

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