# 1959 AHSME Problems/Problem 48

Given the polynomial $a_0x^n+a_1x^{n-1}+\cdots+a_{n-1}x+a_n$, where $n$ is a positive integer or zero, and $a_0$ is a positive integer. The remaining $a$'s are integers or zero. Set $h=n+a_0+|a_1|+|a_2|+\cdots+|a_n|$. [See example 25 for the meaning of $|x|$.] The number of polynomials with $h=3$ is: $\textbf{(A)}\ 3\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 9$

## Solution

We perform casework by the value of $n$, the degree of our polynomial $a_0x^n+a_1x^{n-1}+\cdots+a_{n-1}x+a_n$.

Case $n = 0$: In this case we are forced to set $a_0 = 3$. This contributes $1$ possibility.

Case $n = 1$: In this case we must have $a_0 + |a_1| = 2$, so our polynomial could be $1 + 1x, 0 + 2x, -1 + 1x$. This contributes $3$ possibilities.

Case $n = 2$: In this case we must have $a_0 + |a_1| + |a_2| = 1$. However, because $a_0$ must be positive, it has to be $1$, so our polynomial can only be $0 + 0x + 1x^2$. This contributes $1$ possibility.

Case $n\geq 3$: This case is impossible because $h = n+a_0+|a_1|+|a_2|+\cdots+|a_n|\geq n + a_0\geq 3 + 1 = 4 > 3$, so it contributes $0$ possibilities.

Adding the results from all four cases, we find that there are $1 + 3 + 1 + 0 = 5$ possibilities in total, so our answer is $\boxed{\textbf{(B)}$ (Error compiling LaTeX. ! File ended while scanning use of \boxed.) and we are done.

## See also

 1959 AHSC (Problems • Answer Key • Resources) Preceded byProblem 47 Followed byProblem 49 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 All AHSME Problems and Solutions

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