# There are 4 successive whole numbers that sum to -66. What are these numbers?

Mar 6, 2017

Let the consecutive integers be
$n , n + 1 , n + 2 \mathmr{and} n + 3$

Given condition is
$n + \left(n + 1\right) + \left(n + 2\right) + \left(n + 3\right) = - 66$
$\implies 4 n = - 66 - 1 - 2 - 3$
$\implies 4 n = - 72$
$\implies n = - \frac{72}{4} = - 18$

Consecutive integers are
$- 18 , - 18 + 1 , - 18 + 2 \mathmr{and} - 18 + 3$
$- 18 , - 17 , - 16 \mathmr{and} - 15$

Mar 6, 2017

There is a trick to this type of question.

$- 15 - 16 - 17 - 18 = - 66$

#### Explanation:

Notice that we have negative 66 so I elect to count increasingly negative.

Let a value be $n$

So we have: $\text{ } \left(n\right) + \left(n - 1\right) + \left(n - 2\right) + \left(n - 3\right) = - 66$

$4 n - 6 = - 66$

$4 n = - 60$

$n = - \frac{60}{4} = - 15$

Check: $- 15 - 16 - 17 - 18 = - 66$
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$\textcolor{b l u e}{\text{Would this work if I counted in the direction towards positive. }}$

That is; becoming les negative

$\left(n\right) + \left(n + 1\right) + \left(n + 2\right) + \left(n + 3\right) = - 66$

$4 n + 6 = - 66$

$4 n = - 72$

$n = - 18$

$\left(- 18\right) + \left(- 18 + 1\right) + \left(- 18 + 2\right) + \left(- 18 + 3\right)$

$- 18 - 17 - 16 - 15$