2010 AMC 10B Problem 25

Since $P(x) - a$ has roots $1, 3, 5, 7$, the factor theorem gives
$$P(x) - a = (x-1)(x-3)(x-5)(x-7)\,Q(x)$$
for some polynomial $Q$ with integer coefficients (dividing by a monic integer polynomial keeps integer coefficients).

Plug in the even points, where $P = -a$, so the left side is $-2a$:
- $x = 2$: $(1)(-1)(-3)(-5)Q(2) = -15\,Q(2)$, so $15 \mid 2a$.
- $x = 4$: $(3)(1)(-1)(-3)Q(4) = 9\,Q(4)$, so $9 \mid 2a$.
- $x = 6$: $(5)(3)(1)(-1)Q(6) = -15\,Q(6)$, so $15 \mid 2a$.
- $x = 8$: $(7)(5)(3)(1)Q(8) = 105\,Q(8)$, so $105 \mid 2a$.

Hence $\operatorname{lcm}(15, 9, 105) = 315$ divides $2a$, and since $315$ is odd, $315 \mid a$. So $a \ge 315$.

To see $a = 315$ is attainable we need an integer polynomial $Q$ with $Q(2) = 42$, $Q(4) = -70$, $Q(6) = 42$, $Q(8) = -6$. Write $Q(x) = c_0 + c_1(x-2) + c_2(x-2)(x-4) + c_3(x-2)(x-4)(x-6)$ and solve in order: $c_0 = 42$; $42 + 2c_1 = -70$ gives $c_1 = -56$; $42 - 224 + 8c_2 = 42$ gives $c_2 = 28$; $42 - 336 + 672 + 48c_3 = -6$ gives $c_3 = -8$. All integers, so such a $P$ exists.

The answer is $\boxed{\textbf{(B)}\ 315}$.

Equal values at several points are a signal to subtract that value and factor by the factor theorem; the remaining factor is still an integer polynomial. Evaluating at the other points converts the condition into divisibility by small products, and the lcm of those products is the natural lower bound. A minimum needs both halves: the bound and a construction.

Bound only: choice (A) $105$ fails because $x = 4$ requires $9 \mid 2a$ and $9 \nmid 210$. The remaining choices are all multiples of $315$, so the smallest candidate consistent with the divisibility conditions is (B); the existence check above confirms it.

• Stopping at the lower bound and forgetting to confirm that an integer polynomial Q actually exists for a = 315, or using the product 15915*105 instead of the lcm.
• Forgetting the factor of $2$ in $P(x) - a = -2a$ at the even points, or concluding $315 \mid 2a$ means $a \ge 630$.