lean_workbook_plus_51553

⚠️ Retired — specification defect

The Lean statement below does not encode the problem shown on this page, so its Disproved status carries no information about that problem. Do not import this node or use it as a dependency.

Let's try again from $1111\cdot 10^k+a\equiv 0(2018)$ which implies $102\cdot 10^k+a\equiv 0(1009)$ . For $k=1$ we get $a\equiv -11(1009)$ with no solutions since $0\le a<10$ . For $k=2$ we get $a\equiv 899 (1009)$ with no solutions since $0\le a<100$ . For $k=3$ we find $a\equiv 918(1009)$ . Hence $n=1111918$ is minimal.

Why this node was retired

The posted statement is

theorem lean_workbook_plus_51553  (n a : ℕ)
  (h₀ : 0 < a ∧ a < 10)
  (h₁ : 0 < n ∧ n < 10^7)
  (h₂ : (n * 10^3 + a) % 2018 = 0)
  (h₃ : ∀ m, (m * 10^3 + a) % 2018 = 0 → n ≤ m) :
  n = 1111918 ∧ a = 918   :=  by sorry

The formal statement changes or incorrectly supplies part of the source problem; the accepted note identifies a concrete witness or definitional mechanism.

The recorded counterexample refutes the statement as encoded. It says nothing about the problem shown above, which is a different proposition.

Proposed corrected statement

Find the least integer divisible by 2018 whose decimal expansion begins with 1111; encode N=1111*10^k+a with 0≤a<10^k and minimize N over k≥0 and a. Prove the minimum is 1111918, with k=3,a=918.

Diagnosis and correction from the public Prove2Me statement audit (wamlat/prove2me-errors). The correction is natural-language mathematics and is not Lean-verified — it is a specification for a corrected node, not a drop-in replacement. No corrected replacement node exists yet.