Expert: Scott A Wilson - 2/11/2010

Question
If 6 is subtracted from the largest of three consecutive odd integers, with the result multiplied by 2, the answer is 23 less than the sum of the first and twice the second of the integers. Find the three integers.

Answer
All three odd means we have n, n+2, and n+4.
Subtracting 6 from the largest gives us n + 4 - 6 = n - 2.
Multiplying by 2 gives us 2n - 4.
Since it is 23 less than the other, add 23 to make it so they will be equal.
This gives 2n - 4 + 23 = 2n + 19 (A).

Now the other number we are concerned about is n + 2(n+2) = 3n + 4 (B).

Putting (A) and (B) together, we have 2n + 19 = 3n + .
Subtract 3n from both sides and get -n + 19 = 4
and rewrite it as 19 - n = 4.  It can be seen that n is 15 since 19 - 15 = 4.

That means the numbers are, as n, n+2, and n+4, 15, 17, and 19.

Checking (A), note that 2n + 19 = 2(15) + 19 = 49.
Checking (B), note that 3n + 4 = 3(15) + 4 = 45 + 4 = 49.

As can be seen, that shows the answer is correct.