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

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Find the number of cubic polynomials <math>p(x) = x^3 + ax^2 + bx + c</math>, where <math>a</math>, <math>b</math>, and <math>c</math>
 
Find the number of cubic polynomials <math>p(x) = x^3 + ax^2 + bx + c</math>, where <math>a</math>, <math>b</math>, and <math>c</math>
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are integers in <math>{−20, −10, −18, . . . , 18, 19, 20}</math>, such that there is a unique integer
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are integers in <math>\{ -20, -19, -18, \dots , 18, 19, 20 \}</math>, such that there is a unique integer
−
<math>m \neq 2</math> with <math>p(m) = p(2)</math>.
+
<math>m \neq 2</math> with <math>p(m) = p(2).</math>
−
 
 
−
Find the number of cubic polynomials <math>p(x) = x^3 + ax^2 + bx + c</math>, where <math>a</math>, <math>b</math>, and <math>c</math>
 
−
are integers in <math>{-20, -10, -18, . . . , 18, 19, 20}</math>, such that there is a unique integer
 
−
<math>m \neq 2</math> with <math>p(m) = p(2)</math>.
 
−
 
 
−
Find the number of cubic polynomials <math>p(x) = x^3 + ax^2 + bx + c</math>, where <math>a</math>, <math>b</math>, and <math>c</math>
 
−
are integers in <math>{-20, -10, -18, . . . , 18, 19, 20}</math>, such that there is a unique integer
 
−
<math>m \neq 2</math> with <math>p(m) = p(2)</math>.
 
  
 
==Solution==
 
==Solution==

Revision as of 16:57, 8 February 2023

Find the number of cubic polynomials $p(x) = x^3 + ax^2 + bx + c$, where $a$, $b$, and $c$ are integers in $\{ -20, -19, -18, \dots , 18, 19, 20 \}$, such that there is a unique integer $m \neq 2$ with $p(m) = p(2).$

Solution

It can be easily noticed that $c$ is independent of the condition $P(m) = P(2)$, and can thus safely take all $41$ possible values between $-20$ and $20$.

There are two possible ways for $m\ne2$ to be the only integer satisfying $P(m) = P(2)$: $P$ has a double root at $2$ or a double root at $m$.

Case 1: $P$ has a double root at $2$:

In this case, $\frac{dP}{dx}(2) = 0$, or $12 + 4a + b = 0$. Thus $a$ ranges from $-8$ to $2$. One of these values, $(a,b) = (-6,-12)$ corresponds to a triple root at $2$, which means $m=2$. Thus there are $10$ possible values of $(a,b)$. (It can be verified that $m$ is an integer).

Case 2: $P$ has a double root at $m$:

See the above solution. This yields $8$ possible combinations of $a$ and $b$.

Thus, in total we have $41*18 = \boxed{738}$ combinations of $(a,b,c)$.


-Alex_Z

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