2021 AMC 10A Problem 14

By Vieta, the six roots sum to $10$ and multiply to $16$. Positive integers with product $16 = 2^4$ are all powers of $2$, so the roots come from $\{1, 2, 4, 8, 16\}$.

Distribute four factors of $2$ among six roots so the sum is $10$. Any root of $4$ or more overshoots: the smallest such sums are $4 + 2 + 2 + 1 + 1 + 1 = 11$, $4 + 4 + 1 + 1 + 1 + 1 = 12$, $8 + 2 + 1 + 1 + 1 + 1 = 14$, $16 + 1 + 1 + 1 + 1 + 1 = 21$. So every root is $1$ or $2$. With four $2$s and two $1$s the sum is $8 + 2 = 10$. The polynomial is
$$(z-2)^4 (z-1)^2 = (z^4 - 8z^3 + 24z^2 - 32z + 16)(z^2 - 2z + 1).$$

The $z^3$ coefficient collects the products whose degrees add to $3$:
$$(-8)(1) + (24)(-2) + (-32)(1) = -8 - 48 - 32 = -88.$$

The answer is $\boxed{\textbf{(A)}\ -88}$.

Two Vieta relations (sum and product) plus the integrality condition pin down a multiset of roots that would otherwise need all six coefficients. The bounding step is short because a single large root already overshoots the sum. Once the roots are known, any coefficient is just multiplication, but keep the alternating signs straight: the $z^{n-k}$ coefficient is $(-1)^k$ times the $k$th symmetric sum.

Compute $B$ directly as $-e_3$, the negative of the sum of all triple products of the roots $\{2,2,2,2,1,1\}$: three $2$s give $\binom43 \cdot 8 = 32$; two $2$s and a $1$ give $\binom42 \cdot 2 \cdot 4 = 48$; one $2$ and both $1$s give $4 \cdot 2 = 8$. Total $88$, so $B = -88$.

• Getting the sign wrong: B is the negative of the sum of triple products, so a correct 88 must be reported as -88.
• Forgetting that roots can repeat and searching for six distinct positive integers, which is impossible here.