How do you find three consecutive odd positive integers such that 3 times the sum of all three is 24 more than the product of the first and second integers?

1 Answer
Mar 16, 2016

Answer:

#{1, 3, 5}#

Explanation:

Any odd number may be represented as #2k+1# for some integer #k#. Then, let's represent our odd integers as #2k+1, 2k+3, 2k+5# where #k# is a non-negative integer. From our initial condition, we have that

#(2k+1)(2k+3)+24 = 3((2k+1)+(2k+3)+(2k+5))#

Simplifying both sides, we get

#4k^2+8k+27 = 18k+27#

#=> 4k^2-10k = 0#

#=> 4k(k-5/2) = 0#

#=> k = 0# or #k = 5/2#

But, as we specified that #k# must be an integer, we know that #k!=5/2#. Thus we have #k = 0# and so our odd integers are #{1, 3, 5}#

Checking this, we find that #1*3+24 = 27 = 3(1+3+5)#