2010 AMC 10B Problem 25

This is a "smallest possible" question, so there are two halves: show that $a$ must be at least $315$, then show that $315$ really works.

Step 1: pull out a factor. The polynomial $P(x) - a$ equals $0$ at $x = 1, 3, 5, 7$. By the factor theorem it has the factors $x - 1$, $x - 3$, $x - 5$ and $x - 7$:
$$P(x) - a = (x-1)(x-3)(x-5)(x-7)\,Q(x).$$
Here $Q(x)$ is another polynomial, and its coefficients are also whole numbers (positive, negative or zero). The reason: each factor starts with plain $x$, so dividing $P(x) - a$ by it with long division never creates a fraction.

Part 1: $a$ must be a multiple of $315$.

1. At $x = 2, 4, 6, 8$ we have $P(x) = -a$, so the left side is $-a - a = -2a$. Work out the product of the four factors at each point:

1. $Q$ has whole-number coefficients, so $Q(2)$, $Q(4)$, $Q(6)$, $Q(8)$ are whole numbers. The table then says $2a$ is a multiple of $15$, of $9$ and of $105$.
2. The smallest number that is a multiple of all three: $9 = 3^2$, $15 = 3 \cdot 5$, $105 = 3 \cdot 5 \cdot 7$, so we need $3^2 \cdot 5 \cdot 7 = 315$. So $2a$ is a multiple of $315$.
3. $315$ is odd, so the factor $2$ does not help; $a$ itself must be a multiple of $315$. Since $a > 0$, $a \ge 315$.

Part 2: $a = 315$ works.

6. With $a = 315$, the table tells us what $Q$ must equal: $2a = 630$, so
$$Q(2) = \frac{630}{15} = 42, \quad Q(4) = -\frac{630}{9} = -70, \quad Q(6) = 42, \quad Q(8) = -\frac{630}{105} = -6.$$
We need a polynomial $Q$ with whole-number coefficients that hits these four values.
7. Build it in a form where each new term is zero at the earlier points:
$$Q(x) = c_0 + c_1(x-2) + c_2(x-2)(x-4) + c_3(x-2)(x-4)(x-6).$$
Now find the numbers one at a time.
- $x = 2$: only $c_0$ survives, so $c_0 = 42$.
- $x = 4$: $42 + 2c_1 = -70$, so $c_1 = -56$.
- $x = 6$: $42 + 4(-56) + 4 \cdot 2 \cdot c_2 = 42$, that is $-224 + 8c_2 = 0$, so $c_2 = 28$.
- $x = 8$: $42 + 6(-56) + 6 \cdot 4 \cdot 28 + 6 \cdot 4 \cdot 2 \cdot c_3 = -6$, that is $378 + 48c_3 = -6$, so $c_3 = -8$.
8. All four numbers are whole numbers, so $Q$ has whole-number coefficients, and then $P(x) = 315 + (x-1)(x-3)(x-5)(x-7)Q(x)$ does too. It has $P = 315$ at the odd points and $P = -315$ at the even points.

So $a$ cannot be smaller than $315$, and $315$ is possible.

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

When a polynomial takes the same value at several points, subtract that value and factor: the new polynomial is zero at those points. Plugging in the other points then turns the conditions into "this number must be a multiple of that one", and the smallest number that meets all of them is the lower limit. For a "smallest possible" answer, also build an example that actually reaches it.

If time is short, use Part 1 with the answer choices. Choice (A) $105$ fails: at $x = 4$, $2a = 210$ would have to be a multiple of $9$, and it is not. Every other choice is a multiple of $315$, so the smallest one that could work is $315$. (Part 2 above confirms that it does.)