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Which of the following statements is true?

  1. There are three consecutive integers with sum $2015$
  2. There are four consecutive integers with sum $2015$
  3. There are five consecutive integers with sum $2015$
  4. There are three consecutive integers with product $2015$
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A) There are $3$ consecutive integers with sum $2015$

    $3$ integers are $(x-1),x,(x+1)$

    $(x-1)+x+(x+1) = 2015$

$\implies 3x = 2015$

$\implies x = 2015/3$ 

         $=671.66$

$670.66, 671.66, 672.66$  are not integers.

B) There are $4$ consecutive integers with sum $2015$

    $4$ integers are $(x+1),(x+2),(x+3),(x+4)$

   $(x+1)+(x+2)+(x+3)+(x+4) = 2015$

$\implies 4x +10 = 2015$

$\implies 4x = 2015-10 = 2005$

$\implies x = \dfrac{2005}{4}$

      $=501.25$

$502.25, 503.25, 504.25, 505.25$ are not integers.

C) There are $5$ consecutive integers with sum $2015$

    $5$ integers are $(x-2),(x-1),x,(x+1),(x+2)$

  $ (x-2)+(x-1)+x+(x+1)+(x+2) = 2015$

$\implies 5x = 2015$

$\implies x = 2015/5$

      $=403$

$401, 402, 403, 404, 405$ are integers.

The integers are $(403-2),(403-1),403,(403+1),(403+2) \implies 401,402,403,404,405$

Therefore,sum = $401+402+403+404+405 = 2015$

D) There are $3$ consecutive integers with product $2015$

    $3$ integers are $(x-1),x,(x+1)$

   $(x-1) \times x \times (x+1) = 2015$

$\implies (x^2-x)(x+1) = 2015$

$\implies x^3-x^2+x^2-x = 2015$

$\implies x^3-x = 2015$

$\implies x^3-x-2015 = 0$

it clearly shows that we cannot get integer from this equation.

$\color{green}{∴ \text{Correct option is C) There are 5 consecutive integers with sum 2015}}$

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$let$ $the$ $numbers$ $are$ $x-2 , x-1, x, x+ 1,x+2$

$(x-2) + ( x-1 ) + x + ( x+ 1) +( x+2 ) = 2015$

$5x = 2015$

$x = 2015 / 5$

$x = 403$

numbers are $401,402,403,404,405$

$401 + 402 + 403 + 404 + 405 = 2015$

option C is correct
Answer:

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