How do you find three consecutive odd integers such that twice the sum of the three numbers is 33 or more than the third number?

1 Answer
Mar 21, 2018

Answer:

#5, 7, 9#

Explanation:

Let the three consecutive odd numbers be: #(2n-1), (2n+1), (2n+3)# for some #n in NN#

We are told that twice the sum of these numbers is at least equal to #33 + (2n+3)#

Thus, considering the limiting case, #2(2n-1 +2n+1 +2n+3) = (2n+3)+33#

#2(6n+3) = (2n+3)+33#

#12n+6 = 2n+36#

#10n = 30#

#n=3#

Replacing #n=3# into our sequence of odd numbers #-> {5, 7, 9}#

To check: #5+7+9 = 21#

#2xx21 = 42#

#42-9 =33#