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

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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 ==
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== 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>.
 
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>.
 
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>.
  
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Then, plugging in values of <math>2,4,6,8,</math> we get  
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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(2)-a=(2-1)(2-3)(2-5)(2-7)Q(2) = -15Q(2) = -2a</cmath>
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</center>
 
</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>.
+
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 ==

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
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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

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