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Difference between revisions of "2023 AIME I Problems/Problem 9"

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==Solution==
 
==Solution==
 
===Solution 1===
 
===Solution 1===
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<math>P(x) = x^3 + ax^2 + bx + c</math> is a polynomial with integer coefficients in the range<math>[-20, -19, -18\cdots 18, 19, 20]</math>, inclusive. There is exactly one integer <math>m \neq 2</math> such that <math>P(m) = P(2)</math>. How many possible values are there for the ordered triple <math>(a, b, c)</math>?
 
  
 
Plugging <math>2</math> into <math>P(x)</math>, we get <math>8+4a+2b+c = m^3+am^2+bm+c</math>. We can rewrite into <math>(2-m)(m^2+2m+4+a(2+m)+b)=0</math>, where <math>c</math> can be any value in the range. Since <math>m\neq2, m^2+2m+4+a(2+m)+b</math> must be <math>0</math>. The problem also asks for unique integers, meaning <math>m</math> can only be one value for each polynomial, so the discriminant must be <math>0</math>. <math>m^2+2m+4+a(2+m)+b = m^2+m(2+a)+(2a+b+4)= 0</math>, and <math>(2+a)^2-4(2a+b+4)=0</math>. Rewrite to be <math>a(a-4)=4(b+3)</math>. <math>a</math> must be even for <math>4(b+3)</math> to be an integer. <math>-10<=a<=10</math> because <math>4(20+3) = 92</math>.  There are 11 pairs of <math>(a,b)</math> and 41 integers for <math>c</math>, giving <cmath>41\cdot11 = \boxed{451}</cmath>
 
Plugging <math>2</math> into <math>P(x)</math>, we get <math>8+4a+2b+c = m^3+am^2+bm+c</math>. We can rewrite into <math>(2-m)(m^2+2m+4+a(2+m)+b)=0</math>, where <math>c</math> can be any value in the range. Since <math>m\neq2, m^2+2m+4+a(2+m)+b</math> must be <math>0</math>. The problem also asks for unique integers, meaning <math>m</math> can only be one value for each polynomial, so the discriminant must be <math>0</math>. <math>m^2+2m+4+a(2+m)+b = m^2+m(2+a)+(2a+b+4)= 0</math>, and <math>(2+a)^2-4(2a+b+4)=0</math>. Rewrite to be <math>a(a-4)=4(b+3)</math>. <math>a</math> must be even for <math>4(b+3)</math> to be an integer. <math>-10<=a<=10</math> because <math>4(20+3) = 92</math>.  There are 11 pairs of <math>(a,b)</math> and 41 integers for <math>c</math>, giving <cmath>41\cdot11 = \boxed{451}</cmath>

Revision as of 14:11, 8 February 2023

Problem (Unofficial, please update when official one comes out):

$P(x) = x^3 + ax^2 + bx + c$ is a polynomial with integer coefficients in the range$[-20, -19, -18\cdots 18, 19, 20]$, inclusive. There is exactly one integer $m \neq 2$ such that $P(m) = P(2)$. How many possible values are there for the ordered triple $(a, b, c)$?

Solution

Solution 1

Plugging $2$ into $P(x)$, we get $8+4a+2b+c = m^3+am^2+bm+c$. We can rewrite into $(2-m)(m^2+2m+4+a(2+m)+b)=0$, where $c$ can be any value in the range. Since $m\neq2, m^2+2m+4+a(2+m)+b$ must be $0$. The problem also asks for unique integers, meaning $m$ can only be one value for each polynomial, so the discriminant must be $0$. $m^2+2m+4+a(2+m)+b = m^2+m(2+a)+(2a+b+4)= 0$, and $(2+a)^2-4(2a+b+4)=0$. Rewrite to be $a(a-4)=4(b+3)$. $a$ must be even for $4(b+3)$ to be an integer. $-10<=a<=10$ because $4(20+3) = 92$. There are 11 pairs of $(a,b)$ and 41 integers for $c$, giving \[41\cdot11 = \boxed{451}\]

~chem1kall

Solution 2

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