TITLE: Average of 3 consecutive odd numbers
QUESTION [7 upvotes]: The average of $3$ consecutive odd numbers is $14$ more than one third of the first of these numbers, what is the last of these numbers?
$17/19/15/$data inadequate/none of these

Let three consecutive odd numbers be $a-2,a,a+2$. Their average is $a.$ So, $a=\frac13(a-2)+14\implies a=20.$ But that's not possible as $a$ is odd.
What am I missing?

REPLY [2 votes]: We take $3$ odd numbers $a,b,c$ such that $a= (2n+1), b=(2n+3), c=(2n+5)$ for some integer $n$. From information given, we know $$\frac{(2n+1)+(2n+3)+(2n+5)}{3} = 14 + \frac{1}{3}(2n + 1) \\ \frac{6n+9}{3}=\frac{43+2n}{3} \\ \frac{4n-34}{3} = 0 \\ n = \frac{34}{4} \\ n=8.5$$ Since n is not an integer, the solution cannot exist.