10. Let $N$ be the smallest positive integer whose digits add up to 2012 . What is the first digit of $N+1$ ?
A 2
B 3
C 4
D 5
E 6
Solution: $\mathbf{E}$
If $N$ is the smallest positive integer whose digits add up to 2012, it will have the smallest possible number of digits among numbers whose digits add up to 2012. This means that, as far as possible, each digit should be a 9 . So $N$ will have the form $d \underbrace{999 \ldots 999}_{k}$, where $k$ is a positive integer and $d$ is a non-zero digit. Now, $2012=223 \times 9+5$. So $k=223$ and $d=5$. Hence $N=\underbrace{5999 \ldots 999}_{223}$.
Therefore $N+1=\underbrace{600 \ldots 000}_{223}$. So the first digit of $N+1$ is a 6 .