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

Revision as of 14:18, 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. $-6<=a<=10$ because $4(20+3) = 92$ . There are 9 pairs of $(a,b)$ and 41 integers for $c$ , giving $41\cdot9 = \boxed{369}$

~chem1kall

Solution 2

a=-10 and a=-8 give values for b outside the range and a=-6 results in m=2. Therefore shouldn't the answer be 41*8=328?

@@ Line 6: / Line 6: @@
 ===Solution 1===
-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>-6<=a<=10</math> because <math>4(20+3) = 92</math>.  There are 9 pairs of <math>(a,b)</math> and 41 integers for <math>c</math>, giving <cmath>41\cdot9 = \boxed{369}</cmath>
 ~chem1kall

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