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

Revision as of 14:14, 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

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?

Revision as of 14:11, 8 February 2023 (view source) Chem1kall (talk \| contribs) m (→‎Solution 1) ← Older edit		Revision as of 14:14, 8 February 2023 (view source) Ilikepixd (talk \| contribs) (→‎Solution 2) Newer edit →
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	===Solution 2===		===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?

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