olymmath

HMMT November 2021 H6

  • Theme
  • Answer

Let nn be the answer to this problem. aa and bb are positive integers satisfying

3a+5b≡19( mod n+1)4a+2b≡25( mod n+1)\begin{aligned} & 3 a+5 b \equiv 19 (\bmod n+1) \\ & 4 a+2 b \equiv 25 (\bmod n+1) \end{aligned}

Find 2a+6b2 a+6 b.

Dataset text NuminaMath olympiads. Quoted with source.

Show the answer

96

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 hmnt/2021/h6 · Tag KFWKX5GF9W · Edit this page on GitHub · Keys: j next, k previous