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

(Undo revision 123047 by Jbala (talk))
(Tags: Removed redirect, Undo)
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<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>
  
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== Solution 1 ==
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== Solution ==
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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>.
 +
 
 +
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, plugging in values of <math>2,4,6,8,</math> we get  
 
Then, plugging in values of <math>2,4,6,8,</math> we get  
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<cmath>P(6)-a=(6-1)(6-3)(6-5)(6-7)Q(6) = -15Q(6) = -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>P(8)-a=(8-1)(8-3)(8-5)(8-7)Q(8) = 105Q(8) = -2a</cmath>
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<cmath>-2a=-15Q(2)=9Q(4)=-15Q(6)=105Q(8).</cmath>
 
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Thus, <math>a</math> must be a multiple of <math>\text{lcm}(15,9,15,105)=315</math>.
 
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Now we show that there exists <math>Q(x)</math> such that <math>a=315.</math> We have
 
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<cmath>Q(2)=42, Q(4)=-70, Q(6)=42, Q(8)=-6</cmath>
 
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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>
 
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Therefore, our answer is <math> \boxed{\textbf{(B)}\ 315.} </math>
 
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== Solution 2 (Calculus)==
 
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The evenly-spaced data suggests using [[discrete derivative|discrete derivatives]] to tackle this problem.  First, note that any polynomial of degree <math>n</math>
 
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<center><math>P(x) = a_0 + a_1 x + a_2 x^2 + \ldots + a_n x^n</math></center>
 
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can also be written as
 
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<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>
 
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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>.
 
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The discrete derivative of a function <math>f(x)</math> is the related function <math>\Delta f(x)</math> defined as
 
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<center><math>\Delta f(x) = f(x+1) - f(x)</math>.</center>
 
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With this definition, it's easy to see that for any positive integer <math>k</math> we have
 
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<center><math>\Delta [(x-1)(x-2)\cdots(x-k)] = k(x-1)(x-2)\cdots(x-[k-1])</math>.</center>
 
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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
 
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<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>
 
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We can also calculate the following table of discrete derivatives based on the data points given in the problem statement:
 
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<center>
 
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<table frame='box' rules='all' cellpadding='3'>
 
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<tr><th /><th colspan='8'><math>x</math></th></tr>
 
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<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>
 
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<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>
 
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<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>
 
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<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>
 
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<tr><td colspan='9' align='center'><math>\vdots</math></td></tr>
 
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<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>
 
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</table>
 
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</center>
 
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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
 
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<center>
 
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<math>0! | a</math>,
 
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<math>1! | 2a</math>,
 
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<math>2! | 2^2 a</math>,
 
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<math>3! | 2^3 a</math>,
 
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<math>4! | 2^4 a</math>,
 
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<math>5! | 2^5 a</math>,
 
  
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<math>6! | 2^6 a</math>, and
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<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>.
  
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<math>7! | 2^7 a</math>.
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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
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</center>
 
  
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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>.
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<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>
  
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== See also ==
 
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{{AMC12 box|year=2010|num-b=20|num-a=22|ab=B}}
 
  
 
[[Category:Intermediate Algebra Problems]]
 
[[Category:Intermediate Algebra Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 15:18, 17 June 2020

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

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, 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

\[P(x) = (x-1)(x-3)(x-5)(x-7)((-8x + 60)(x-2)(x-6)+42) + 315\] \[= -8 x^7+252 x^6-3248 x^5+22050 x^4-84392 x^3+179928 x^2-194592 x+80325\] The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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