2000 AMC 10 Problem 14

1. Turn the averages into sums. An average of $k$ scores is a whole number exactly when their sum is a multiple of $k$. So:
- the first $2$ scores add to an even number,
- the first $3$ scores add to a multiple of $3$,
- the first $4$ scores add to a multiple of $4$,
- all $5$ add to $71 + 76 + 80 + 82 + 91 = 400$.
2. Look at the last score. The first four add to a multiple of $4$, and $400$ is a multiple of $4$. So the last score, which is $400$ minus the first four, is also a multiple of $4$. Among the scores, only $76$ and $80$ are multiples of $4$.
3. Remainders when dividing by $3$. We will need these:

4. Which three can come first? The first three must add to a multiple of $3$, so their remainders must add to $0$, $3$ or $6$. Trying the combinations, the only choice is the three scores with remainder $1$: $76, 82, 91$, whose sum is $249 = 3 \cdot 83$. (Two 2s and a 1 add to $5$; one 2 and two 1s add to $4$.)
5. So $76, 82, 91$ are entered first, and $71$ and $80$ are the last two. From step 2 the last score is a multiple of $4$, so it is $80$.
6. Check the whole order $76, 82, 91, 71, 80$:
- $76$: average $76$.
- $76 + 82 = 158$, average $79$.
- $158 + 91 = 249$, average $83$.
- $249 + 71 = 320$, average $80$.
- $320 + 80 = 400$, average $80$.

Every average is a whole number.

The answer is $\boxed{\textbf{(C)}\ 80}$.

"The average is a whole number" is really a statement about dividing evenly. Each step of the entry gives one divisibility condition, and the conditions for $3$ and $4$ were enough to pin down the order. When you see "always a whole number average," switch from averages to sums right away.

Go straight to step 4. The remainders after dividing by $3$ are $2, 1, 2, 1, 1$, and the only three that add to a multiple of $3$ are the three 1s, so $76, 82, 91$ come first. Their sum $249$ is odd, and the first four must add to a multiple of $4$, so the fourth score must be odd too: it is $71$ ($249 + 71 = 320$). That leaves $80$ for last.