1959 AMC 12 Problem 48

Attempt Problem 48 of the 1959 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1959 AMC 12 solutions, or check the answer key.

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48.

Given the polynomial a0xn+a1xn−1+⋯+an−1x+an, \begin{aligned} &a_0x^n+a_1x^{n-1}+\cdots\\ &\qquad{}+a_{n-1}x+a_n, \end{aligned} where nn is a positive integer or zero, and a0a_0 is a positive integer. The remaining aa’s are integers or zero. Set h=n+a0+∣a1∣+∣a2∣+⋯+∣an∣. \begin{aligned} h={}&n+a_0+|a_1|+|a_2|\\ &{}+\cdots+|a_n|. \end{aligned} [See example 2525 for the meaning of ∣x∣.|x|.] The number of polynomials with h=3h=3 is:
33
55
66
77
99
Answer: B
Concepts:polynomialcasework
Small hint:
Since n+a0≤3n+a_0\le3 and a0≥1,a_0\ge1, consider only n=0,1,2n=0,1,2
Big hint:
For each degree, count the integer coefficient tuples with the required sum of absolute values
Solution:
We count by degree. If n=0,n=0, then a0=3,a_0=3, giving one polynomial. If n=1,n=1, then a0+∣a1∣=2. a_0+|a_1|=2. This gives a0=2,a1=0a_0=2,a_1=0 or a0=1,a1=±1,a_0=1,a_1=\pm1, for three polynomials. If n=2,n=2, then a0=1a_0=1 and a1=a2=0,a_1=a_2=0, giving one more. No higher degree is possible. The total is 1+3+1=5.1+3+1=5.
Thus, the correct answer is B.
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