Maths Olympiad Prep

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Algebra Difficulty 4.9 AIME Find the answer

4. It is known that there exist integers aa, bb, and cc such that the equation (xa)(x1998)+1=(x+b)(x+c)(x-a) \cdot (x-1998) + 1 = (x+b)(x+c) holds for any real number xx. Then the value of 2a+b+c|2a + b + c| is

A number or a short expression. Spacing and $ signs are ignored.

Solution

4.24.2

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