Letters for Digits: Puzzles and Logic

TL;DR

Letters for Digits puzzles, also known as cryptarithmetic, involve replacing letters with unique digits to make a math equation true. You'll use logic, number properties, and some trial-and-error to crack these codes. The goal is to find a unique digit for each letter, ensuring the arithmetic works out correctly.

1. The Mental Model

Think of these puzzles as a secret code where each letter hides a different number. Your job is to break that code. You're like a detective, using clues from the math problem to figure out which number each letter stands for.

2. The Core Material

Letters for Digits puzzles present an arithmetic problem where letters stand in for digits. The main rules are:
1. Each letter represents a unique digit (0-9).
2. No number can start with zero (e.g., if 'A' is the first letter of a multi-digit number, A cannot be 0).
3. The arithmetic (addition, subtraction, multiplication) must be correct.

These puzzles are solved through a combination of logical deduction and strategic trial-and-error.

### Deductive Reasoning: Starting Strong

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The most powerful way to begin is by looking for immediate clues.

• Look for carries: In addition, if you add two single-digit numbers, the sum can't be more than 18 (9+9). If you add three, it's max 27 (9+9+9). This tells you a lot about possible carry-overs to the next column. For example, in AB + CD = EF, if A and C are single digits and E is a single digit, then E could be A + C or A + C + carry-in. If the result has one more digit than the addends, the leftmost digit must be a carry-over, often '1'.
  • Example: SEND + MORE = MONEY. M in MONEY must be '1' because it's a carry-over from S + M (or S + M + carry) in the hundreds column.
• Analyze the leftmost digits: As seen above, the first digit of a number cannot be zero. This is a crucial constraint.
• Identify simple operations:
  • A + A = B: This means B must be an even number.
  • A + B = A: This implies B must be 0 (or B + carry = 0, meaning B is 9 with a carry-in of 1).
  • A + 0 = A
  • A * 0 = 0
  • A * 1 = A
  • A * 5 will always end in 0 or 5.
• Parity (Even/Odd): Sometimes the sum or product of numbers can tell you if a digit is even or odd.

### Strategic Trial-and-Error

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Once you've exhausted direct deductions, you'll need to start testing possibilities.
1. Prioritize letters with few options: If a letter can only be 0 or 1, try those first.
2. Make temporary assignments: Jot down your guesses.
3. Check for conflicts: Does your guess for one letter make another part of the puzzle impossible? Does it violate the unique digit rule?
4. Backtrack: If a guess leads to a contradiction, undo it and try the next possibility.

graph TD
    A["Start Puzzle"] --> B{"Any obvious deductions? (e.g., carries, leading zeros)"}
    B -- "Yes" --> C["Apply deductions (assign digits, narrow possibilities)"]
    C --> D{"All letters assigned?"}
    D -- "No" --> E{"Are there letters with limited options remaining?"}
    E -- "Yes" --> F["Select a letter; try one of its remaining possible digits"]
    F --> G{"Check if this assignment creates contradictions (e.g., duplicate digits, impossible arithmetic)"}
    G -- "No conflicts" --> C
    G -- "Conflicts found" --> H["Backtrack: Undo assignment; try next possible digit for that letter"]
    H --> F
    E -- "No (many options for all remaining)" --> I["Choose a letter; make an educated guess"]
    I --> G
    D -- "Yes" --> J{"Does the arithmetic equation hold true with all assigned digits?"}
    J -- "Yes" --> K["Solution Found!"]
    J -- "No" --> H

3. Worked Example

Let's solve the classic:

SEND + MORE = MONEY

1. M must be 1: In the addition of S + M (thousands column), MONEY has one more digit than SEND or MORE. This means there must have been a carry-over, and M is the carried digit. So, M = 1.

2. S must be 8 or 9: Since M is 1, and S + M (plus any carry from E+O) results in MO (which is 1O), S + 1 (+ carry) must be at least 10. If S were 7 or less, even with a carry of 1, S+1+1 would be 9 or less, making M not 1. So S is 8 or 9.
   Also, S + M (or S + M + carry_EON) leads to MO. Since M is 1, and S + 1 needs to generate a carry of 1, S must be 9 to make 9 + 1 = 10 (if there's no carry-in from E+O). Or S is 8 if there is a carry-in (8+1+1=10). We'll assume S=9 for now, meaning O=0 (from 9+1=10, so O in MONEY is 0 and we carry 1 to the next column). If S=8, then O=0 as well (8+1+1=10). This means O=0 is a very strong candidate.

3. O must be 0: From step 2, if S=9 and M=1, then S + M + (carry from E+O) produces MO. 9 + 1 + (carry) means 10 + (carry). For the result to be MONEY starting with 10..., O must be 0. So, O = 0.

4. E and N: Look at the column E + O = N (with a potential carry-in from N+R and a potential carry-out). We know O = 0. So E + 0 + (carry from N+R) results in N.
   If there's no carry-in from N+R, then E = N. But letters must be unique. So there must be a carry-in from N+R to E+O. This means E + 0 + 1 = N, or E + 1 = N. This implies that N is one greater than E.

5. N and R: N + R = E (or E+10 if there's a carry-out to the E+O column). There was a carry-out to E+O (from step 4). So N + R = E + 10.

6. D and E: D + E = Y (or Y+10 if there's a carry to the N+R column). There was a carry to the N+R column (from step 5). So D + E = Y + 10.

Let's list what we have:
M = 1
O = 0
S = 9 (from 9+1=10, O=0, and a carry of 1 to next column)
E + 1 = N (and we carried 1 to the next column)
N + R = E + 10 (since we carried 1 to the E+O column)
D + E = Y + 10 (since we carried 1 to the N+R column)

Remaining digits: 2, 3, 4, 5, 6, 7, 8.
We know E+1=N. Possible pairs for (E, N): (2,3), (3,4), (4,5), (5,6), (6,7), (7,8).
Try E=5. Then N=6.
If E=5, N=6:
N + R = E + 10 becomes 6 + R = 5 + 10, so 6 + R = 15. This means R = 9.
But S is already 9! This combination for E, N, R doesn't work.

Let's retry E+1=N pairs.
Consider N+R=E+10. Since N = E+1, substitute: (E+1) + R = E + 10.
This simplifies to 1 + R = 10, so R = 9.
We already assigned S=9. So R cannot be 9. This means our assumption S=9 (meaning S+1=10) was wrong.
This indicates there must have been a carry-in to the S+M column!

Let's re-evaluate S:
S + M + carry_from_EON = MO
S + 1 + 1 (carry from E+O=N) = 10 + O
S + 2 = 10 + O
Since O=0, then S + 2 = 10, so S = 8.

So, revised assignments:
M = 1
O = 0
S = 8 (because there was a carry-in from E+O=N column)

Now we have E + 1 = N (still true, and carry to S column is 1)
And N + R = E + 10 (still true, and carry to E column is 1)
And D + E = Y + 10 (still true, and carry to N column is