Difference between revisions of "2023 AIME I Problems/Problem 9"
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− | a=-10 and a=-8 give values for b outside the range and a=-6 results in m= | + | 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 14:14, 8 February 2023
Contents
[hide]Problem (Unofficial, please update when official one comes out):
is a polynomial with integer coefficients in the range, inclusive. There is exactly one integer such that . How many possible values are there for the ordered triple ?
Solution
Solution 1
Plugging into , we get . We can rewrite into , where can be any value in the range. Since must be . The problem also asks for unique integers, meaning can only be one value for each polynomial, so the discriminant must be . , and . Rewrite to be . must be even for to be an integer. because . There are 11 pairs of and 41 integers for , giving
~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?