# How do you find four consecutive integers whose sum in twice the cube of 5?

Mar 13, 2018

$61 , 62 , 63 , 64$

#### Explanation:

You can treat the four consecutive integers as $\left(x\right) , \left(x + 1\right) , \left(x + 2\right) ,$ and $\left(x + 3\right)$.

You set up the equation:
$\left(x\right) + \left(x + 1\right) + \left(x + 2\right) + \left(x + 3\right) = 2 \cdot \left({5}^{3}\right)$

Solve the equation:
$4 x + 6 = 250$
$4 x = 244$
$x = 61$

Now, you can plug in $x$ back into your original four consecutive integers.

$\left(61\right) , \left(61 + 1\right) , \left(61 + 2\right) , \left(61 + 3\right)$,
which gives you:
$61 , 62 , 63 , 64$

Mar 13, 2018

61, 62, 63, 64

#### Explanation:

$\left(x\right) + \left(x + 1\right) + \left(x + 2\right) + \left(x + 3\right) = 2 \left({5}^{3}\right)$

$\left(x\right) + \left(x + 1\right) + \left(x + 2\right) + \left(x + 3\right) = 2 \left(125\right)$

$\left(x\right) + \left(x + 1\right) + \left(x + 2\right) + \left(x + 3\right) = 250$

So now we can just take out the parenthesis on the left side of the equation and simplify

$\left(x\right) + \left(x + 1\right) + \left(x + 2\right) + \left(x + 3\right) = 250$

$4 x + 6 = 250$

$4 x = 244$

$x = 61$