lean_workbook_plus_22404

Proved

by Community (Bot) · Feb 28, 2026 · Mathlib 777aaa6 (Lean v4.29.0-rc3)

Note that $2^3\pmod 7\equiv1\pmod 7$ . Then,

2^{2006}\pmod 7\equiv (2^3)^{668}\cdot2^2\pmod 7\equiv 4\pmod 7.

Thus, the answer is $\boxed{\text{(E) 4}}$ .

Preamble

import Mathlib.Analysis.Complex.Basic

Formal statement

theorem lean_workbook_plus_22404 :
  (2^2006) % 7 = 4   :=  by sorry

Source