https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Star32&feedformat=atom AoPS Wiki - User contributions [en] 2021-01-26T13:37:12Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_12B_Problems/Problem_22&diff=130220 2018 AMC 12B Problems/Problem 22 2020-08-02T04:42:21Z <p>Star32: /* Solution 3 (BASH) */</p> <hr /> <div>== Problem ==<br /> <br /> Consider polynomials &lt;math&gt;P(x)&lt;/math&gt; of degree at most &lt;math&gt;3&lt;/math&gt;, each of whose coefficients is an element of &lt;math&gt;\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}&lt;/math&gt;. How many such polynomials satisfy &lt;math&gt;P(-1) = -9&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 110 \qquad \textbf{(B) } 143 \qquad \textbf{(C) } 165 \qquad \textbf{(D) } 220 \qquad \textbf{(E) } 286 &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> Suppose our polynomial is equal to<br /> &lt;cmath&gt;ax^3+bx^2+cx+d&lt;/cmath&gt;Then we are given that<br /> &lt;cmath&gt;-9=b+d-a-c.&lt;/cmath&gt;If we let &lt;math&gt;-a=a'-9, -c=c'-9&lt;/math&gt; then we have<br /> &lt;cmath&gt;9=a'+c'+b+d.&lt;/cmath&gt; This way all four variables are within 0 and 9. The number of solutions to this equation is simply &lt;math&gt;\binom{12}{3}=220&lt;/math&gt; by stars and bars, so our answer is &lt;math&gt;\boxed{\textbf{D}.}&lt;/math&gt;<br /> <br /> == Solution 2 ==<br /> Suppose our polynomial is equal to<br /> &lt;cmath&gt;ax^3+bx^2+cx+d&lt;/cmath&gt;Then we are given that<br /> &lt;cmath&gt;9=b+d-a-c.&lt;/cmath&gt;Then the polynomials &lt;cmath&gt;cx^3+bx^2+ax+d&lt;/cmath&gt;, &lt;cmath&gt;ax^3+dx^2+cx+b,&lt;/cmath&gt; &lt;cmath&gt;cx^3+dx^2+ax+b&lt;/cmath&gt;also have &lt;cmath&gt;b+d-a-c=-9&lt;/cmath&gt; when &lt;cmath&gt;x=-1.&lt;/cmath&gt; So the number of solutions must be divisible by 4. So the answer must be &lt;math&gt;\boxed{\textbf{D}.}&lt;/math&gt;<br /> <br /> <br /> == Solution 3 ==<br /> As before, &lt;math&gt;-9=-A+B-C+D&lt;/math&gt;. This is &lt;math&gt;9=(A+B)-(C+D)&lt;/math&gt;. Rephrased, how many two sums of integers from 0-9 have a difference of 9. Make a chart of pairs between these two sets:<br /> &lt;cmath&gt;{0,1,2,3,4,5,6,7,8,9}&lt;/cmath&gt;<br /> &lt;cmath&gt;{9,10,11,12,13,14,15,16,17,18}&lt;/cmath&gt;<br /> Observe how there is one way to sum 2 numbers to 0 and two ways to 1, 3 ways to 2, and so on. At 9, there are 10 possible ways. Recall that only integers between 0-9 are valid. Now observe how there is 1 way to to sum to 18 in this fashion (9+9), 2 ways to sum to 17, and so forth again (to optionally prove that this pattern holds, apply stars and bars up to 9 and notice the symmetry).<br /> <br /> The answer then is the number of ways to write each component of each pair. This is &lt;math&gt;(1*10)+(2*9)+(3*8) ...&lt;/math&gt; or, since it's symmetrical between sum of 4 and 5, &lt;math&gt;[2\sum\limits_{i=1}^{5} i(11-i)]&lt;/math&gt;. Use summation rules to finally get &lt;math&gt;\boxed{\textbf{D)}220}&lt;/math&gt;.<br /> <br /> ~BJHHar<br /> <br /> ==See Also==<br /> <br /> {{AMC12 box|year=2018|ab=B|num-b=21|num-a=23}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> [[Category:Intermediate Algebra Problems]]</div> Star32