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

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Answer 1 · 2017-03-06T17:02:00

Let the consecutive integers be
$n, n+1, n+2 and n+3$

Given condition is
$n+( n+1)+ (n+2)+( n+3)=-66$
$=>4n=-66-1-2-3$
$=>4n=-72$
$=>n=-72/4=-18$

Consecutive integers are
$-18, -18+1, -18+2 and -18+3$
$-18, -17, -16 and -15$

Answer 2 · 2017-03-06T17:11:01

There is a trick to this type of question.

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

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

Let a value be $n$

So we have: $" "(n)+(n-1)+(n-2)+(n-3)=-66$

$4n-6=-66$

$4n=-60$

$n=-60/4=-15$

Check: $-15-16-17-18= -66$
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$color(blue)("Would this work if I counted in the direction towards positive. ")$

That is; becoming les negative

$(n)+(n+1)+(n+2)+(n+3)=-66$

$4n+6=-66$

$4n=-72$

$n=-18$

$(-18)+(-18+1)+(-18+2)+(-18+3)$

$-18-17-16-15$