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

Latest revision as of 01:53, 20 November 2024

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 $\boxed{\textbf{(B)}\ 315}$ .

@@ Line 1: / Line 1: @@
-== Problem 21 ==
+== Problem ==
 Let <math>a > 0</math>, and let <math>P(x)</math> be a polynomial with integer coefficients such that
 <center>
-<math>P(1) = P(3) = P(5) = P(7) = a</math>, and<br/>
+<math>P(1) = P(3) = P(5) = P(7) = a</math>,
+and <br >
 <math>P(2) = P(4) = P(6) = P(8) = -a</math>.
 </center>
@@ Line 12: / Line 13: @@
 == Solution 1 ==
-There must be some polynomial <math>Q(x)</math> such that <math>P(x)-a=(x-1)(x-3)(x-5)(x-7)Q(x)</math>
+We observe that because <math>P(1) = P(3) = P(5) = P(7) = a</math>, if we define a new polynomial <math>R(x)</math> such that <math>R(x) = P(x) - a</math>, <math>R(x)</math> has roots when <math>P(x) = a</math>; namely, when <math>x=1,3,5,7</math>.
-Then, plugging in values of <math>2,4,6,8,</math> we get
+Thus since <math>R(x)</math> has roots when <math>x=1,3,5,7</math>, we can factor the product <math>(x-1)(x-3)(x-5)(x-7)</math> out of <math>R(x)</math> to obtain a new polynomial <math>Q(x)</math> such that <math>(x-1)(x-3)(x-5)(x-7)(Q(x)) = R(x) = P(x) - a</math>.
+Then, noting that <math>P(2)-a=-a-a=-2a</math> etc., and plugging in values of <math>2,4,6,8,</math> we get
 <cmath>P(2)-a=(2-1)(2-3)(2-5)(2-7)Q(2) = -15Q(2) = -2a</cmath>
@@ Line 20: / Line 23: @@
 <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>.
-Now we show that there exists <math>Q(x)</math> such that <math>a=315.</math> We have
+<math>-2a=-15Q(2)=9Q(4)=-15Q(6)=105Q(8).</math>
-<cmath>Q(2)=42, Q(4)=-70, Q(6)=42, Q(8)=-6</cmath>
+Thus, the least value of <math>a</math> must be the <math>\text{lcm}(15,9,15,105)</math>.
-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>
+Solving, we receive <math>315</math>, so our answer is <math> \boxed{\textbf{(B)}\ 315} </math>.
+To complete the solution, we can let <math>a = 315</math>, and then try to find <math>Q(x)</math>. We know from the above calculation that <math>Q(2)=42, Q(4)=-70, Q(6)=42</math>, and <math>Q(8)=-6</math>. Then we can let <math>Q(x) = T(x)(x-2)(x-6)+42</math>, getting <math>T(4)=28, T(8)=-4</math>. Let <math>T(x)=L(x)(x-8)-4</math>, then <math>L(4)=-8</math>. Therefore, it is possible to choose <math>T(x) = -8(x-8)-4 = -8x + 60</math>, so the goal is accomplished. As a reference, the polynomial we get is
-Therefore, our answer is <math> \boxed{\textbf{(B)}\ 315.} </math>
+<cmath>P(x) = (x-1)(x-3)(x-5)(x-7)((-8x + 60)(x-2)(x-6)+42) + 315</cmath>
+<cmath> = -8 x^7+252 x^6-3248 x^5+22050 x^4-84392 x^3+179928 x^2-194592 x+80325 </cmath>
-== Solution 2 (Calculus)==
+== Solution 2 ==
 The evenly-spaced data suggests using [[discrete derivative|discrete derivatives]] to tackle this problem.  First, note that any polynomial of degree <math>n</math>
 <center><math>P(x) = a_0 + a_1 x + a_2 x^2 + \ldots + a_n x^n</math></center>
 can also be written as
-<center><math>P(x) = b_0 + b_1 (x-1) + b_2 (x-1)(x-2) + \ldots + b_n (x-1)(x-2) \cdots (x-n)</math>.</center>
+<center><math>P(x) = b_0 + b_1 (x-1) + b_2 (x-1)(x-2) + \ldots + b_n (x-1)(x-1) \cdots (x-n)</math>.</center>
-Moreover, the coefficients <math>a_i</math> are integers for <math>i=0, 1, 2, \ldots n</math> iff the coefficients <math>b_i</math> are integers for <math>i=0, 1, 2, \ldots n</math>.  This latter form is convenient for calculating discrete derivatives of <math>P(x)</math>.
+Moreover, the coefficients <math>a_i</math> are integers for <math>i=1, 2, \ldots n</math> iff the coefficients <math>b_i</math> are integers for <math>i=1, 2, \ldots n</math>.  This latter form is convenient for calculating discrete derivatives of <math>P(x)</math>.
 The discrete derivative of a function <math>f(x)</math> is the related function <math>\Delta f(x)</math> defined as
 <center><math>\Delta f(x) = f(x+1) - f(x)</math>.</center>
 With this definition, it's easy to see that for any positive integer <math>k</math> we have
 <center><math>\Delta [(x-1)(x-2)\cdots(x-k)] = k(x-1)(x-2)\cdots(x-[k-1])</math>.</center>
 This in turn allows us to use successive discrete derivatives evaluated at <math>x=1</math> to calculate all of the coefficients <math>b_i</math> using
 <center><math>P(1)=b_0</math>, <math>\Delta P(1) = b_1</math>, <math>\Delta^2 P(1) = 2 b_2</math>, <math>\ldots</math>, <math>\Delta^7 P(1) = 7! b_7</math>.</center>
 We can also calculate the following table of discrete derivatives based on the data points given in the problem statement:
 <center>
 <table frame='box' rules='all' cellpadding='3'>
 <tr><th /><th colspan='8'><math>x</math></th></tr>
 <tr><td align='right'></td><td align='center'><math>1</math></td><td align='center'><math>2</math></td><td align='center'><math>3</math></td><td align='center'><math>4</math></td><td align='center'><math>5</math></td><td align='center'><math>6</math></td><td align='center'><math>7</math></td><td align='center'><math>8</math></td></tr>
 <tr><td align='right'><math>P(x)</math></td><td align='right'><math>a</math></td><td align='right'><math>-a</math></td><td align='right'><math>a</math></td><td align='right'><math>-a</math></td><td align='right'><math>a</math></td><td align='right'><math>-a</math></td><td align='right'><math>a</math></td><td align='right'><math>-a</math></td></tr>
 <tr><td align='right'><math>\Delta P(x)</math></td><td align='right'><math>-2a</math></td><td align='right'><math>2a</math></td><td align='right'><math>-2a</math></td><td align='right'><math>2a</math></td><td align='right'><math>-2a</math></td><td align='right'><math>2a</math></td><td align='right'><math>-2a</math></td><td /></tr>
 <tr><td align='right'><math>\Delta^2 P(x)</math></td><td align='right'><math>4a</math></td><td align='right'><math>-4a</math></td><td align='right'><math>4a</math></td><td align='right'><math>-4a</math></td><td align='right'><math>4a</math></td><td align='right'><math>-4a</math></td><td /><td /></tr>
 <tr><td colspan='9' align='center'><math>\vdots</math></td></tr>
 <tr><td align='right'><math>\Delta^7 P(x)</math></td><td align='right'><math>-2^7 a</math></td><td /><td /><td /><td /><td /><td /><td /></tr>
 </table>
 </center>
 Thus we can read down the column for <math>x=1</math> to find that <math>k! b_k = (-2)^k a</math> for <math>k = 0, 1, \ldots, 7</math>.  Interestingly, even if we choose <math>P(x)</math> to have degree greater than <math>7</math>, the <math>8</math> coefficients of lowest order always satisfy these conditions.  Moreover, it's straightforward to show that the <math>P(x)</math> of degree <math>7</math> with <math>b_k</math> satisfying these conditions will fit the data given in the problem statement.  Thus, to find minimal necessary and sufficient conditions on the value of <math>a</math>, we need only consider these <math>8</math> equations.  As a result, <math>P(x)</math> with integer coefficients fitting the given data exists iff <math>k!</math> divides <math>2^k a</math> for <math>k = 0, 1, \ldots, 7</math>.  In other words, it's necessary and sufficient that
 <center>
 <math>0! | a</math>,
 <math>1! | 2a</math>,
 <math>2! | 2^2 a</math>,
 <math>3! | 2^3 a</math>,
 <math>4! | 2^4 a</math>,
 <math>5! | 2^5 a</math>,
 <math>6! | 2^6 a</math>, and
 <math>7! | 2^7 a</math>.
 </center>
-The last condition holds iff <math>7 \cdot 3 \cdot 5 \cdot 3 = 315</math> divides evenly into <math>a</math>.  Since such <math>a</math> will also satisfy the first <math>7</math> conditions, our answer is <math> \boxed{\textbf{(B)}\ 315} </math>.
+The last condition holds if <math>7 \cdot 3 \cdot 5 \cdot 3 = 315</math> divides evenly into <math>a</math>.  Since such <math>a</math> will also satisfy the first <math>7</math> conditions, our answer is <math> \boxed{\textbf{(B)}\ 315} </math>.
 == See also ==
-{{AMC12 box|year=2010|num-b=20|num-a=22|ab=B}}
+{{AMC12 box|year=2010|ab=B|num-b=20|num-a=22}}
+{{AMC10 box|ab=B|year=2010|after=Last Problem|num-b=24}}
 [[Category:Intermediate Algebra Problems]]
 {{MAA Notice}}

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

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

Latest revision as of 01:53, 20 November 2024

Contents

Problem

Solution 1

Solution 2

See also