2000 AMC 10 Problem 14

An integer average after $k$ entries means the first $k$ scores sum to a multiple of $k$. The five scores total $400$.

Mod 4: the first four scores sum to a multiple of $4$, and $400$ is a multiple of $4$, so the last score is a multiple of $4$: it is $76$ or $80$.

Mod 3: the residues of $71, 76, 80, 82, 91$ are $2, 1, 2, 1, 1$. The first three scores sum to a multiple of $3$, so the fourth score's residue equals the residue of the first four scores' sum.

• If the last score is $76$, the first four sum to $324$, a multiple of $3$, so the fourth score would have residue $0$. No score does. Rejected.
• If the last score is $80$, the first four sum to $320 \equiv 2 \pmod 3$, so the fourth score has residue $2$: it must be $71$.

The remaining scores $76, 82, 91$ come first; they sum to $249$, a multiple of $3$, and $76 + 82 = 158$ is even, so the order $76, 82, 91, 71, 80$ (or with the first two swapped) satisfies every condition.

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

"Average is an integer" is a divisibility statement in disguise. Working backwards from the known total, each divisibility condition constrains one more position; two residues ($\bmod 4$, then $\bmod 3$) were enough here. Tagging note: the statistics language is superficial; the work is modular arithmetic.

• Checking only the mod-4 condition (last score is 76 or 80) and guessing, instead of also using the mod-3 condition on the fourth score.
• Forgetting to verify that a full valid order exists after the last two positions are forced.