Difference between revisions of "2010 AMC 12B Problems/Problem 21"

Revision as of 10:03, 15 August 2017

Problem 21

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

There must be some polynomial $Q(x)$ such that $P(x)-a=(x-1)(x-3)(x-5)(x-7)Q(x)$

Then, 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, $a$ must be a multiple of $\text{lcm}(15,9,15,105)=315$ .

Now we show that there exists $Q(x)$ such that $a=315.$ We have $Q(2)=42, Q(4)=-70, Q(6)=42, Q(8)=-6$ Thus, $Q(x)=R(x)(x-2)(x-6)+42$ for some $R(x).$ From here it is clear that $Q(x)$ exists, since we can take $R(x)=-8x+60.$ $\blacksquare$

Therefore, our answer is $\boxed{\textbf{(B)}\ 315}$ .

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 $i=0, 1, 2, \ldots n$ iff the coefficients $b_i$ are integers for $i=0, 1, 2, \ldots n$ . 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 iff $7 \cdot 3 \cdot 5 \cdot 3 = 315$ divides evenly into $a$ . Since such $a$ will also satisfy the first $7$ conditions, our answer is $\boxed{\textbf{(B)}\ 315}$ .

@@ Line 16: / Line 16: @@
 Then, plugging in values of <math>2,4,6,8,</math> we get
-<math>P(2)-a=(2-1)(2-3)(2-5)(2-7)Q(2) = -15Q(2) = -2a</math>
+<cmath>P(2)-a=(2-1)(2-3)(2-5)(2-7)Q(2) = -15Q(2) = -2a</cmath>
+<cmath>P(4)-a=(4-1)(4-3)(4-5)(4-7)Q(4) = 9Q(4) = -2a</cmath>
+<cmath>P(6)-a=(6-1)(6-3)(6-5)(6-7)Q(6) = -15Q(6) = -2a</cmath>
+<cmath>P(8)-a=(8-1)(8-3)(8-5)(8-7)Q(8) = 105Q(8) = -2a</cmath>
+<cmath>-2a=-15Q(2)=9Q(4)=-15Q(6)=105Q(8).</cmath>
+Thus, <math>a</math> must be a multiple of <math>\text{lcm}(15,9,15,105)=315</math>.
-<math>P(4)-a=(4-1)(4-3)(4-5)(4-7)Q(4) = 9Q(4) = -2a</math>
+Now we show that there exists <math>Q(x)</math> such that <math>a=315.</math> We have
+<cmath>Q(2)=42, Q(4)=-70, Q(6)=42, Q(8)=-6</cmath>
+Thus, <math>Q(x)=R(x)(x-2)(x-6)+42</math> for some <math>R(x).</math> From here it is clear that <math>Q(x)</math> exists, since we can take <math>R(x)=-8x+60.</math> <math>\blacksquare</math>
-<math>P(6)-a=(6-1)(6-3)(6-5)(6-7)Q(6) = -15Q(6) = -2a</math>
+Therefore, our answer is <math> \boxed{\textbf{(B)}\ 315} </math>.
-<math>P(8)-a=(8-1)(8-3)(8-5)(8-7)Q(8) = 105Q(8) = -2a</math>
-<math>-2a=-15Q(2)=9Q(4)=-15Q(6)=105Q(8).</math>
-Thus, the least value of <math>a</math> must be the <math>\text{lcm}(15,9,15,105)</math>.
-Solving, we receive <math>315</math>, so our answer is <math> \boxed{\textbf{(B)}\ 315} </math>.
-(This solution appears to be incomplete in that it only shows that <math>a</math> being a multiple of <math>315</math> is a <em>necessary</em> condition for <math>P(x)</math> to have integer coefficients.  However, it's not clear at this point that <math>a=315</math> is <em>sufficient</em> to guarantee that <math>P(x)</math> has only integer coefficients. For example, at this point how do we know that for <math>a=315</math> the equations above for <math>Q(2)</math>, <math>Q(4)</math>, <math>Q(6)</math>, and <math>Q(8)</math> don't require that, say, the second coefficient of <math>Q(x)</math> is <math>1/97</math>?)
 == Solution 2 ==

2010 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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Difference between revisions of "2010 AMC 12B Problems/Problem 21"

Revision as of 10:03, 15 August 2017

Contents

Problem 21

Solution 1

Solution 2

See also