In the cryptarithmetic puzzle SEND + MORE = MONEY, where each letter…

Concept: In a cryptarithmetic (alphametic) puzzle, every distinct letter stands for a unique digit from 0-9, letters that begin a number cannot be 0, and the letters must satisfy ordinary column-by-column addition with carries, exactly like adding numbers by hand.

Application: Align SEND, MORE, and MONEY by place value and work column by column, tracking exactly which carry each column must produce:

1. Adding two 4-digit numbers can carry at most 1 into a new fifth digit, so the leading digit of MONEY, M, must be 1, and this carry of 1 is what feeds into the thousands column.

2. Thousands column: S + M, plus any carry from the hundreds column, must equal O with exactly this carry of 1 going out. Substituting M = 1 gives S + carry(hundreds) = O + 9, leaving two candidates: S = 9 with carry(hundreds) = 0, or S = 8 with carry(hundreds) = 1 (O cannot be 1, since M already is). The second candidate needs the hundreds column itself to carry 1 while O = 0, but the hundreds-column equation only allows that if it forces N = 0 — and O is already 0, so two letters cannot share a digit, ruling this branch out. That leaves S = 9, O = 0, with no carry coming in from the hundreds column.

3. Hundreds column: E + O, plus any carry from the tens column, must equal N (mod 10). With O = 0, this reduces to N = E + carry(tens). If that carry were 0, N would equal E, which is impossible since every letter is a distinct digit — so the tens column must carry 1 into the hundreds column, giving N = E + 1.

4. Tens column: N + R, plus any carry from the units column, must equal E with a carry of 1 out (from the hundreds column, step 3). Substituting N = E + 1 gives R + carry(units) = 9; since R = 9 would clash with S = 9, the units column must carry 1, giving R = 8.

5. Units column: D + E must equal Y with a carry of 1 out (from step 4), i.e., D = Y + 10 - E. Testing the digits still unused ({2, 3, 4, 5, 6, 7} for E, with N = E + 1 also needing to stay in that set) shows only E = 5 lets D and Y both land on still-available digits: D = 7, Y = 2 (with N = 6); every other choice of E forces D or Y outside the digits that remain.

6. This gives S = 9, E = 5, N = 6, D = 7, M = 1, O = 0, R = 8, Y = 2, so SEND = 9567 and MORE = 1085.

7. Adding them: 9567 + 1085 = 10652, so MONEY = 10652 — confirming M = 1, O = 0, N = 6, E = 5, Y = 2.

Cross-check: Re-adding column by column independently — units 7 + 5 = 12 (write 2, carry 1); tens 6 + 8 + 1 = 15 (write 5, carry 1); hundreds 5 + 0 + 1 = 6; thousands 9 + 1 = 10 — reproduces 10652, confirming the digit assignment.

So, M + O + N + E + Y = 1 + 0 + 6 + 5 + 2 = 14.