2016 AMC 10A Problem 9

Recall that the sum of the first $N$ numbers is $\dfrac{N(N + 1)}{2}.$ We want to find $N$ such that $$\dfrac{N(N + 1)}{2} = 2016.$$ Cross-multiplying and simplifying gives us $$N^2 + N - 4032 = 0.$$ Factoring gives us $$(N - 63)(N + 64) = 0.$$ We want the positive value so $N = 63.$ Adding together the digits gives us $9.$ Thus, the correct answer is D .