HMMT 2023 A10
- Algebra and Number Theory
- Answer
Let ζ=e2πi/99 and ω=e2πi/101. The polynomial
x9999+a9998x9998+⋯+a1x+a0
has roots ζm+ωn for all pairs of integers (m,n) with 0≤m<99 and 0≤n<101. Compute a9799+a9800+⋯+a9998.
Dataset text NuminaMath olympiads. Quoted with source.
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14849-\frac{9999}{200}\binom{200}{99}
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