2002 AMC 10A Problem 21

This asks for the greatest possible largest number, so there are two halves: show $15$ (or more) is impossible, then give a list that works with $14$.

What the conditions say.

1. Mean $8$ with eight numbers: the total is $8 \times 8 = 64$.
2. Range $8$: smallest $=$ largest $- 8$.
3. Median $8$: with eight numbers in order, the 4th and 5th average to $8$.
4. Unique mode $8$: the number $8$ appears more often than any other number.

Part 1: the largest cannot be $15$ or more.

1. Suppose the largest is $15$. Then the smallest is $15 - 8 = 7$, so all eight numbers are at least $7$. The other seven numbers add to $64 - 15 = 49$, and seven numbers that are each at least $7$ add to at least $49$. So they must all be exactly $7$: the list is $7, 7, 7, 7, 7, 7, 7, 15$. Its median and mode are $7$, not $8$. So $15$ fails.

1. Suppose the largest is $16$ or more. Then the smallest is at least $8$, so the total is at least $7 \times 8 + 16 = 72$, which is more than $64$. That fails too.

Part 2: the largest can be $14$.

7. With largest $14$, the smallest is $6$. The other seven numbers must add to $64 - 14 = 50$. Put $8$s in the middle for the median, and use as many $6$s as possible to keep the total down. Three $6$s and four $8$s give $18 + 32 = 50$. The list is
$$6, 6, 6, 8, 8, 8, 8, 14.$$

8. Check every condition:
- total $= 18 + 32 + 14 = 64$, so the mean is $8$;
- the 4th and 5th numbers are $8$ and $8$, so the median is $8$;
- $8$ appears four times and $6$ only three times, so $8$ is the only mode;
- range $= 14 - 6 = 8$.

The largest number can be $14$ but not more.

The answer is $\boxed{\textbf{(D)}\ 14}$.

With statistics conditions, the mean fixes the total, so making one number bigger forces the others to shrink. The range ties the smallest number to the largest, and the unique mode limits how many copies of a small number you can use. Test the largest choice, find which rule it breaks, then build a list for the next choice and check every rule.