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#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#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#