1959 AHSME Problems/Problem 48
Given the polynomial , where is a positive integer or zero, and is a positive integer. The remaining 's are integers or zero. Set . [See example 25 for the meaning of .] The number of polynomials with is:
Solution
We perform casework by the value of , the degree of our polynomial .
Case : In this case we are forced to set . This contributes possibility.
Case : In this case we must have , so our polynomial could be . This contributes possibilities.
Case : In this case we must have . However, because must be positive, it has to be , so our polynomial can only be . This contributes possibility.
Case : This case is impossible because , so it contributes possibilities.
Adding the results from all four cases, we find that there are possibilities in total, so our answer is .
See also
1959 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 47 |
Followed by Problem 49 | |
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