The houses on one side of a road are numbered using consecutive even numbers 
The houses on one side of a road are numbered using consecutive even numbers. The sum of the numbers of all the houses in that row is 170. If there are at least 6 houses in that row and $a$ is the number of the sixth house, then
  (A) $2 ≤ a ≤ 6$
  (B) $8 ≤ a ≤ 12$
  (C) $14 ≤ a ≤ 20$
  (D) $22 ≤ a ≤ 30$     

I couldn't figure out a proper way of solving this, I tried guess-and-check but even that didn't really work. Can somebody give me a hint as to how I can start this?

Suppose there are $k$ houses numbered $c,c+2,c+4,\dots,c+2(k-1)$. The sum of house numbers is
$$kc+k(k-1)=k(c+k-1)=170=2\cdot5\cdot17$$
$k$ is a factor of 170, and the constraint $k\ge6$ limits it to the following numbers:
$$10,17,34,85,170$$
Since the sum of increments $0+2+4+\cdots+2(k-1)=k(k-1)$ must be less than 170, yet $17×16=272>170$, we must have $k=10$, from which we get $c=8$. The sixth house number $a$ is then $c+10=18$ and the answer is (c).
