Consider a country with 15 cities. For 1 ≤ j ≤ 15, let $x_j$ denote the number of roads that lead out of city j to other cities in the country. Adding up $x_j$ for each city j results in a number that is at least 135. Prove that there is a city that must have at least 10 roads leading out of it.
My current logic is as follows. I am not sure if this is correct.
For n cities and n-1 roads/city you can draw n*(n-1) roads between them for remaining 15 - n cities you can draw (15 - n) * (15 - n - 1) = (15 - n)(14 - n) roads between them.
$n(n-1) + (15-n)(14-n) = 135$
$2n^2 - 30n + 75 = 0$
$n = 3.1699$ OR $n = 11.830$
Remember $n$ is the number of cities, thus $n-1$ is the number of roads and there must be at least one city with 10.830 roads thus there must be one city with at least 10 roads leading out of it.