olymmath

HMMT 2023 A10

  • Algebra and Number Theory
  • Answer

Let ζ=e2πi/99\zeta=e^{2 \pi i / 99} and ω=e2πi/101\omega=e^{2 \pi i / 101}. The polynomial

x9999+a9998x9998+⋯+a1x+a0x^{9999}+a_{9998} x^{9998}+\cdots+a_{1} x+a_{0}

has roots ζm+ωn\zeta^{m}+\omega^{n} for all pairs of integers (m,n)(m, n) with 0≤m<990 \leq m<99 and 0≤n<1010 \leq n<101. Compute a9799+a9800+⋯+a9998a_{9799}+a_{9800}+\cdots+a_{9998}.

Dataset text NuminaMath olympiads. Quoted with source.

Show the answer

14849-\frac{9999}{200}\binom{200}{99}

Solutions

No solution is on olymmath yet. Official and human solutions come first when they arrive. AI solutions come after them, with a label.

Add a solution

Similar problems

These problems have the most words in common with this statement.

Locator hmmt/2023/a10 · Tag QWVNFAY2PN · Edit this page on GitHub · Keys: j next, k previous