2001 AMC 10 Problem 1

1. Check the order first. The numbers $n, n+3, n+4, \ldots, n+15$ go up from left to right, so the list is already sorted.
2. There are $9$ numbers. The median of $9$ sorted numbers is the one in the middle, with $4$ on each side. That is the fifth number, $n + 6$.
3. The median is $10$, so $n + 6 = 10$ and $n = 4$.
4. Now write out the actual list using $n = 4$:
$$4,\ 7,\ 8,\ 9,\ 10,\ 12,\ 14,\ 16,\ 19.$$
5. Add them: $4 + 7 + 8 + 9 + 10 + 12 + 14 + 16 + 19 = 99$.
6. The mean is the total divided by how many numbers there are: $99 \div 9 = 11$.

The answer is $\boxed{\textbf{(E)}\ 11}$.

Every number in the list is "$n$ plus something," so once the median tells you $n$, the whole list is known. A useful shortcut: if you add the same amount to every number in a list, the mean and the median both go up by that same amount. Here the extra amounts $0, 3, 4, 5, 6, 8, 10, 12, 15$ have median $6$ and mean $7$, so the mean of the list is always $1$ more than its median.

Use that shortcut directly. The extra amounts add up to $0+3+4+5+6+8+10+12+15 = 63$, and $63 \div 9 = 7$, so the mean is $n + 7$. The median is $n + 6$. The mean is therefore $1$ more than the median: $10 + 1 = 11$. You never need to find $n$.

The nine entries are listed in increasing order, so the median is the fifth one, $n + 6$. Setting $n + 6 = 10$ gives $n = 4$.

For the mean, add the offsets: $0 + 3 + 4 + 5 + 6 + 8 + 10 + 12 + 15 = 63$. The total of the list is $9n + 63$, so the mean is
$$\frac{9n + 63}{9} = n + 7 = 4 + 7 = 11.$$

The answer is $\boxed{\textbf{(E)}\ 11}$.