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).

Problem

T I M E
I S
A
G R E A T
H E A L E R

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

E + S + A + T = R + 10i  (1)
M + I + A + i = E + 10j  (2)
I + E + j = L + 10k  (3)
T + R + k = A + 10ℓ  (4)
G + ℓ = E + 10m  (5)

• Observe:

i ≤ 3 (E + S + A + T ≤ 9 + 8 + 7 + 6 = 30)
j ≤ 2  (M + I + A + i ≤ 9 + 8 + 7 + 3 = 27)
k ≤ 1  (I + E + j ≤ 9 + 8 + 2 = 19)
ℓ ≤ 1  (T + R + k ≤ 9 + 8 + 1 = 18)
m ≤ 1 (G + ℓ ≤ 9 + 1 = 10)

So H = 1 E = 0 G = 9 ℓ = 1 (since H = m ≤ 1 and H ≠ 0 by Rule 2, and so G + ℓ = 9 + 1 = 10 = E + 10)

• i = 1 since … 9 = 0 + 2 + 3 + 4 ≤ E + S + A + T ≤ 0 + 6 + 7 + 8 = 21 and E + S + A + T = R + 10i; also R ≠ 9 = G, R ≠ 0 = E, R ≠ 1 = H; so E + S + A + T ∉ {9, 20, 21}. So i = 1.

• k = 0 since … 10 = 8 + 0 + 2 ≥ I + E + j = L + 10k; also L ≠ 0 = E, so I + E + j ≤ 9; So k = 0.

• A ∈ {2, 3, 4, 5} (since 15 = 7 + 8 + k ≥ T + R + k = A + 10, A ≠ 0 = E, A ≠ 1 = H)

• A ≠ 5 since suppose A = 5 then by (1) and (4)…

S + T + 5 = R + 10
T + R = 15

So T ≥ 7 (since R ≤ 8); so T ∈ {7, 8} (since T ≠ 9 = G). Suppose T = 7 then R = 8 and S = R + 10 − (T + 5) = 6

and so M, I ∈ {2, 3, 4} (2,3,4 are the only digits left);

and M + I + 6 = M + I + A + k = 10j, so j = 1 and M = I = 2, contradicting

Rule 1. So T ≠ 7.

Suppose T = 8 then R = 7 and S = R + 10 − (T + 5) = 4 and so M, I ∈ {2, 3, 6} (2,3,6 are the only digits left); and M + I + 6 = M + I + A + k = 10j, which is impossible. So T ≠ 8.

So A ≠ 5.

• A ≠ 4 since suppose A = 4 then by (1) and (4)…

S + T + 4 = R + 10
T + R = 14

So T ≥ 6 (since R ≤ 8); so T ∈ {6, 7, 8} (since T ≠ 9 = G). Suppose T = 6 then R = 8 and S = R + 10 − (T + 4) = 6 = T, contradicting Rule 1;

So T ≠ 6. Suppose T = 7 then R = 7 = T, contradicting Rule 1;

So T ≠ 7. Suppose T = 8 then R = 6 and S = R + 10 − (T + 4) = 4 = A, contradicting Rule 1;

So T ≠ 8. So A ≠ 4.

• A ≠ 2 since suppose A = 2 then by (1) and (4)…

S + T + 2 = R + 10  (6)
T + R = 12

0 ≥ S − 8 = R − T (by (6)), so T > R (since S ≤ 8); so T ≥ 6 (since T + R = 12); so T ∈ {6, 7, 8} (since T ≠ 9 = G). Suppose T = 6 then R = 6 = T, contradicting Rule 1;

So T ≠ 6. Suppose T = 7 then R = 5 = T and S = R + 10 − (T + 2) = 6,

and so M, I, L ∈ {3, 4, 8} (3,4,8 are the only digits left);

and M + I + 3 = M + I + A + k = 10j, so M + I = 7, M, I ∈ {3, 4}, L = 8; so 6 = 4 + 0 + 2 ≤ I + E + j = L = 8, which is impossible. So T ≠ 7.

Suppose T = 8 then R = 4 and S = R + 10 − (T + 2) = 4 = R, contradicting Rule 1;

So T ≠ 8. So A ≠ 2.

So A = 3

• So, by (1) and (4)…

S + T + 3 = R + 10
T + R = 13

• T ∈ {5, 6, 7, 8} (since T ≠ 9 = G and T = 13 − R ≥ 13 − 8 = 5)

• T ≠ 5 since suppose T = 5 then R = 8 and S = R + 10 − (T + 3) = 10 > 9.

• T ≠ 7 since suppose T = 7 then R = 6 and S = R + 10 − (T + 3) = 6 = R, contradicting Rule 1.

• T ≠ 8 since suppose T = 8 then R = 5 and S = R + 10 − (T + 3) = 4 and … M, I, L ∈ {2, 6, 7} (2,6,7 are the only digits left); and 12 ≤ M + I + 4 ≤ 17; so M + I + 4 = M + I + A + k = 10j cannot be satisfied. So T ≠ 8.

So T = 6 R = 7 S = 8

• So M, I, L ∈ {2, 4, 5}

• L = 5 j = 1 since …

M + I + 4 = M + I + A + k = 10j, so j = 1, M + I = 6, M, I ∈ {2, 4}; So L = 5.

• I = 4 (since I + 1 = I + E + j = L = 5)

• M = 2 (2 is the only digit left)

• So finally we get (and check)

6 4 2 0
4 8

3

9 7 0 3 6
1 0 3 5 0 7