Alphametics

May 3, 2001

The rules of alphametics 1. Each letter represents a distinct digit in {0, 1, …, 9}.

2. The left-most digit of a number is never 0.

3. An alphametic represents a sum (of decimal numbers).

The real problem It’s not too hard to find a solution. It is much harder to prove that a solution found is the only solution … you need to consider all the possibilities and eliminate them until only one possibility remains … and check that that possibility is a solution.

An example

\begin{aligned}& & & A \\ & T & A & N \\ & C & A & N \\ L & A & S & T \end{aligned} T AN is odd.

• First let k, ℓ, m be the carries, so that

A + 2N = T + 10k  (1)
2A + k = S + 10ℓ  (2)
T + C + ℓ = A + 10m  (3)

• Observe:

k ≤ 2  (A + 2N ≤ 8 + 2.9 = 26)
ℓ ≤ 2  (2A + k ≤ 2.9 + 2 = 20)
m ≤ 1 (T + C + ℓ ≤ 9 + 8 + 2 = 19)

So L = 1 (since L = m ≤ 1 and L ≠ 0 by Rule 2)

• N ∈ {3, 5, 7, 9} (T AN is odd and N ≠ L = 1 by Rule 1)

• T > A since … T + C + ℓ = A + 10, ℓ ≤ 2 So T = A + 10 − (C + ℓ) Suppose C + ℓ > 10 then C = 9, ℓ = 2 but then 2A + k = S + 20 and A = 9 = C, which contradicts Rule 1. So C + ℓ ≯ 10.

Suppose C + ℓ = 10 then T = A, which contradicts Rule 1.

So C + ℓ ≠ 10. So C + ℓ < 10. So T > A.

• A ∉ {0, 1, 9} (A ≠ 9 since T > A, A ≠ 0 by Rule 2, A ≠ L = 1)

• N ≠ 3 since suppose N = 3 then … A + 6 = T + 10k (by (1)) and k ≤ 1 (since A ≤ 8). Suppose k = 0 then A + 6 = T;

and so A ≤ 3 (since T ≤ 9); also A ∉ {0, 1}, A ≠ N = 3; so A = 2, T = 8; so 4 = 2A + k = S + 10ℓ (by (2)), so S = 4, ℓ = 0; so 8 + C = T + C = A + 10 = 12 (by (3)); so C = 4 = S, contradicting Rule 1. So k ≠ 0.

Suppose k = 1 then A = T + 4, contradicting T > A.

So k ≠ 1. So N ≠ 3.

• N ≠ 5 since suppose N = 5 then … A + 10 = T + 10k (by (1)) and so k = 1 and A = T, contradicting Rule 1. So N ≠ 5.

• N ≠ 9 since suppose N = 9 then … A + 18 = T + 10k (by (1)) and so k ∈ {1, 2}. Suppose k = 1 then A + 8 = T;

and so A ≤ 1 (since T ≤ 9), but A ∉ {0, 1}. So k ≠ 1.

Suppose k = 2 then A = T + 2 (contradicting T > A).

So k ≠ 2. So N ≠ 9.

So N = 7

• k ∈ {1, 2} (since A + 14 = A + 2N = T + 10k by (1)).

• k = 1, A + 4 = T since … Suppose k = 2 then A + 14 = T + 20, by (1), contradicting T > A. So k = 1 and A + 14 = T + 10. So A + 4 = T.

• A ∈ {2, 4, 5} since … A ∉ {0, 1} and A ≤ 5 (since A + 4 = T ≤ 9); also A ≠ 3 since otherwise T = 7 = N, contradicting Rule 1.

• A ≠ 2 since suppose A = 2 then… 6 = A + 4 = T and 5 = 2A + k = S + 10ℓ, so S = 5, ℓ = 0; so 6 + C = T + C + ℓ = A + 10 = 12, by (3); so C = 6 = T, contradicting Rule 1. So A ≠ 2.

• A ≠ 5 since suppose A = 5 then… 11 = 2A + k = S + 10ℓ, so S = 1 = L, contradicting Rule 1. So A ≠ 5.

So A = 4 T = 8

• So S = 9 ℓ = 0 (since 9 = 2A + k = S + 10ℓ, by (2))

• So C = 6 (since 8 + C = T + C + ℓ = A + 10 = 14, by (3))

• So finally we get (and check)

4 8 4 7 6 4 7 1 4 9 8 847 is odd.