In the alphametic puzzle NINA = AGAIN − SING, each letter stands for one fixed…

In an alphametic (cryptarithmetic) puzzle, every distinct letter represents exactly one digit from 0 to 9 throughout the sum, the same letter always stands for the same digit, and the leading letter of any number in the sum cannot be 0. Such a puzzle is solved by writing the words as a column addition and resolving the columns one at a time — from the units place upward — using the carry each column produces to pin down the next unknown digit.

Rearranging NINA = AGAIN − SING as an addition gives NINA + SING = AGAIN:

N I N A

+ S I N G

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A G A I N

1. Ten-thousands column: NINA and SING are both 4-digit numbers, so their sum can carry over at most one extra digit. That new leading digit is A, so A = 1, and the carry out of the thousands column is 1.

2. Hundreds column: I + I, plus the carry coming in from the tens column, must reproduce A = 1 (with a carry of 0 into the thousands column) or 11 (with a carry of 1). Since 2I is always even, the carry into the hundreds column must be 1 in either case, which gives I = 0 (with no further carry out of hundreds) or I = 5 (with a carry of 1 out of hundreds).

3. Testing I = 5: the tens column N + N + (carry-in) = 15 forces N = 7, and the units column A + G = N + 10 then needs 1 + G = 17, so G = 16 — not a valid single digit. So I = 5 cannot hold.

4. Hence I = 0. The tens column N + N (with the carry of 1 required into the hundreds column) equals 10, so N = 5.

5. Units column: A + G = N gives 1 + G = 5, so G = 4.

6. Thousands column: N + S (with the carry of 1 required into the ten-thousands column) equals G + 10, so 5 + S = 14, giving S = 9.

Substituting back: NINA = 5051 and SING = 9054, and 5051 + 9054 = 14105, which matches AGAIN with A = 1, G = 4, I = 0, N = 5. All five digits (0, 1, 4, 5, 9) are distinct, and none of the leading letters N, S, or A is 0, so the assignment satisfies every condition of the puzzle.

Therefore, S + I + N + G = 9 + 0 + 5 + 4 = 18.