2010 AMC 12B Problems/Problem 21

Problem

Let $a > 0$ , and let $P(x)$ be a polynomial with integer coefficients such that

$P(1) = P(3) = P(5) = P(7) = a$ , and
$P(2) = P(4) = P(6) = P(8) = -a$ .

What is the smallest possible value of $a$ ?

$\textbf{(A)}\ 105 \qquad \textbf{(B)}\ 315 \qquad \textbf{(C)}\ 945 \qquad \textbf{(D)}\ 7! \qquad \textbf{(E)}\ 8!$

Solution 1

We observe that because $P(1) = P(3) = P(5) = P(7) = a$ , if we define a new polynomial $R(x)$ such that $R(x) = P(x) - a$ , $R(x)$ has roots when $P(x) = a$ ; namely, when $x=1,3,5,7$ .

Thus since $R(x)$ has roots when $x=1,3,5,7$ , we can factor the product $(x-1)(x-3)(x-5)(x-7)$ out of $R(x)$ to obtain a new polynomial $Q(x)$ such that $(x-1)(x-3)(x-5)(x-7)(Q(x)) = R(x) = P(x) - a$ .

Then, noting that $P(2)-a=-a-a=-2a$ etc., and plugging in values of $2,4,6,8,$ we get

$P(2)-a=(2-1)(2-3)(2-5)(2-7)Q(2) = -15Q(2) = -2a$ $P(4)-a=(4-1)(4-3)(4-5)(4-7)Q(4) = 9Q(4) = -2a$ $P(6)-a=(6-1)(6-3)(6-5)(6-7)Q(6) = -15Q(6) = -2a$ $P(8)-a=(8-1)(8-3)(8-5)(8-7)Q(8) = 105Q(8) = -2a$

$-2a=-15Q(2)=9Q(4)=-15Q(6)=105Q(8).$ Thus, the least value of $a$ must be the $\text{lcm}(15,9,15,105)$ . Solving, we receive $315$ , so our answer is $\boxed{\textbf{(B)}\ 315}$ .

To complete the solution, we can let $a = 315$ , and then try to find $Q(x)$ . We know from the above calculation that $Q(2)=42, Q(4)=-70, Q(6)=42$ , and $Q(8)=-6$ . Then we can let $Q(x) = T(x)(x-2)(x-6)+42$ , getting $T(4)=28, T(8)=-4$ . Let $T(x)=L(x)(x-8)-4$ , then $L(4)=-8$ . Therefore, it is possible to choose $T(x) = -8(x-8)-4 = -8x + 60$ , so the goal is accomplished. As a reference, the polynomial we get is

$P(x) = (x-1)(x-3)(x-5)(x-7)((-8x + 60)(x-2)(x-6)+42) + 315$ $= -8 x^7+252 x^6-3248 x^5+22050 x^4-84392 x^3+179928 x^2-194592 x+80325$

Solution 2

The evenly-spaced data suggests using discrete derivatives to tackle this problem. First, note that any polynomial of degree $n$

$P(x) = a_0 + a_1 x + a_2 x^2 + \ldots + a_n x^n$

can also be written as

$P(x) = b_0 + b_1 (x-1) + b_2 (x-1)(x-2) + \ldots + b_n (x-1)(x-1) \cdots (x-n)$ .

Moreover, the coefficients $a_i$ are integers for $i=1, 2, \ldots n$ iff the coefficients $b_i$ are integers for $i=1, 2, \ldots n$ . This latter form is convenient for calculating discrete derivatives of $P(x)$ .

The discrete derivative of a function $f(x)$ is the related function $\Delta f(x)$ defined as

$\Delta f(x) = f(x+1) - f(x)$ .

With this definition, it's easy to see that for any positive integer $k$ we have

$\Delta [(x-1)(x-2)\cdots(x-k)] = k(x-1)(x-2)\cdots(x-[k-1])$ .

This in turn allows us to use successive discrete derivatives evaluated at $x=1$ to calculate all of the coefficients $b_i$ using

$P(1)=b_0$ , $\Delta P(1) = b_1$ , $\Delta^2 P(1) = 2 b_2$ , $\ldots$ , $\Delta^7 P(1) = 7! b_7$ .

We can also calculate the following table of discrete derivatives based on the data points given in the problem statement:

	$x$
	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$
$P(x)$	$a$	$-a$	$a$	$-a$	$a$	$-a$	$a$	$-a$
$\Delta P(x)$	$-2a$	$2a$	$-2a$	$2a$	$-2a$	$2a$	$-2a$
$\Delta^2 P(x)$	$4a$	$-4a$	$4a$	$-4a$	$4a$	$-4a$
$\vdots$
$\Delta^7 P(x)$	$-2^7 a$

Thus we can read down the column for $x=1$ to find that $k! b_k = (-2)^k a$ for $k = 0, 1, \ldots, 7$ . Interestingly, even if we choose $P(x)$ to have degree greater than $7$ , the $8$ coefficients of lowest order always satisfy these conditions. Moreover, it's straightforward to show that the $P(x)$ of degree $7$ with $b_k$ satisfying these conditions will fit the data given in the problem statement. Thus, to find minimal necessary and sufficient conditions on the value of $a$ , we need only consider these $8$ equations. As a result, $P(x)$ with integer coefficients fitting the given data exists iff $k!$ divides $2^k a$ for $k = 0, 1, \ldots, 7$ . In other words, it's necessary and sufficient that

$0! | a$ ,

$1! | 2a$ ,

$2! | 2^2 a$ ,

$3! | 2^3 a$ ,

$4! | 2^4 a$ ,

$5! | 2^5 a$ ,

$6! | 2^6 a$ , and

$7! | 2^7 a$ .

The last condition holds if $7 \cdot 3 \cdot 5 \cdot 3 = 315$ divides evenly into $a$ . Since such $a$ will also satisfy the first $7$ conditions, our answer is $\boxed{\textbf{(B)}\ 315}$ .

2010 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

2010 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 24	Followed by Last Problem
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

Page

Toolbox

Search

2010 AMC 12B Problems/Problem 21

Contents

Problem

Solution 1

Solution 2

See also