13) Let $N$ be the largest of the natural numbers $n$ that satisfy the inequality

$$
\frac{n}{n+1}<\frac{6024}{6027}
$$

What is the sum of the digits of $N$?
(A) 6, (B) 7,
(C) 8,
(D) 9,
(E) 10
13. The answer is $(\mathbf{D})$.

We observe that $\frac{6024}{6027}=\frac{2008}{2009}$; therefore, the inequality that must be satisfied can be written in the form

$$
\frac{2008}{2009}>\frac{n}{n+1}
$$

We then see that the inequality is true for $n=2007$ and false for $n=2008$. On the other hand, the quantity

$$
\frac{n}{n+1}=1-\frac{1}{n+1}
$$

increases as $n$ increases, since the fraction $\frac{1}{n+1}$ decreases as $n$ increases. Therefore, the inequality is true for $n=2007$ and for all values $n \leq 2007$, while it is false for $n \geq 2008$. We conclude that $N=2007$ and the sum of its digits is 9.