How do you find three consecutive odd integers such that four times the middle integer is twice the sum of the other two integers?

1 Answer
May 28, 2015

Nice question.

Any three consecutive odd integers have this property. (As do any three consecutive even integers.)

To solve this, let the consecutive odd integers be #n-2#, #n#, and #n+2#.

(We'll throw out any non-odd solutions.)

We want #4(n) = 2[(n-2)+(n+2)]#

But #(n-2)+(n+2) = 2n#, so we want:

#4(n) = 2[(n-2)+(n+2)] = 2[2n] = 4n#

We want: #4n=4n#.

But for any number, #4n=4n#, so any number (every number) is a solution to #4(n) = 2[(n-2)+(n+2)]#.

The question asks only for odd solutions, so:

Any (every) three consecutive odd integers will solve the problem.

Additional Notes

We could have used #n, n+2, "and " n+4# as our 3 numbers.

Or we could make sure they are odd by making the numbers:

#2k+1#, #2k+3#, and #2k+5# in whihc case, we solve:

#4(2k+3) = 2(2k+1+2k+5)# and we get:

#8k+12 = 2(4k+6) = 8k+12#

Again, every number #k# gives a solution.