Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(Created page with '== Problem 21 == 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<b…')
 
(→‎Solution 1)
 
(25 intermediate revisions by 16 users not shown)
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
 
Let <math>a > 0</math>, and let <math>P(x)</math> be a polynomial with integer coefficients such that
  
 
<center>
 
<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>.
 
<math>P(2) = P(4) = P(6) = P(8) = -a</math>.
 
</center>
 
</center>
Line 11: Line 12:
 
<math>\textbf{(A)}\ 105 \qquad \textbf{(B)}\ 315 \qquad \textbf{(C)}\ 945 \qquad \textbf{(D)}\ 7! \qquad \textbf{(E)}\ 8!</math>
 
<math>\textbf{(A)}\ 105 \qquad \textbf{(B)}\ 315 \qquad \textbf{(C)}\ 945 \qquad \textbf{(D)}\ 7! \qquad \textbf{(E)}\ 8!</math>
  
−
== Solution ==
+
== Solution 1 ==
 +
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>.
 +
 
 +
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>
 +
<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>
 +
 
 +
<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>.
 +
 
 +
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
 +
 
 +
<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 ==
 +
 +
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-1) \cdots (x-n)</math>.</center>
 +
 +
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 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 ==
 
== See also ==
−
{{AMC12 box|year=2010|num-b=12|num-a=14|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}}

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

See also

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12B_Problems/Problem_21&oldid=235129"