Each letter in the following addition represents a unique digit from 0 to 9:…

This is a cryptarithmetic (alphametic) puzzle: each distinct letter stands for a unique digit from 0 to 9, and the leading letter of any number can never be 0. Solve it by adding column by column from the units digit outward, using each column's digit sum and its carry into the next column to pin down one letter at a time, then verify by substituting all the digits back into the original sum.

1. Write the addition aligned by place value: the units digits of FORTY, TEN and TEN must add to the units digit of SIXTY, and so on leftward through the tens, hundreds, thousands and ten-thousands columns, carrying into the next column whenever a column sum reaches 10 or more.

2. Units column: Y + N + N = Y with some carry into the tens column, so 2N must be a multiple of 10. That gives N = 0 (no carry) or N = 5 (carry 1 into the tens column).

3. Test N = 5: the tens column becomes T + E + E + 1 = T with a carry, which forces 2E + 1 to be a multiple of 10 — impossible, since 2E + 1 is always odd. So N = 5 is ruled out, leaving N = 0 with no carry out of the units column.

4. Tens column (carry 0 in): T + E + E = T with some carry, so 2E must be a multiple of 10, giving E = 0 or E = 5. Since N already uses 0, E = 5, carrying 1 into the hundreds column.

5. Since FORTY contributes no digit above the ten-thousands place while SIXTY's leading digit is S, the ten-thousands column is simply F plus a carry-in from the thousands column, with no further carry out (the sum has only five digits). As F and S must be different digits, that carry-in must be exactly 1, giving F + 1 = S.

6. Thousands column: only O (from FORTY) feeds this column, so O + carry-in-from-hundreds must equal I with a carry of 1 into the ten-thousands column, i.e. O + carry-in = I + 10. Since 0 and 5 are already used (by N and E), a hundreds-column carry-in of 1 would force I = 0, which is taken — so the carry-in must be 2, giving O + 2 = I + 10, i.e. O = I + 8. The only digit pair satisfying this with 0 and 5 excluded is O = 9, I = 1.

7. Hundreds column: R + T + T + 1 (the carry from the tens column) must supply exactly a carry of 2 into the thousands column, so R + 2T + 1 = X + 20, i.e. R + 2T − X = 19. With 0, 1, 5 and 9 already used, R, T and X must come from the remaining digits {2, 3, 4, 6, 7, 8}. Since R and X can differ by at most 6 within this set, 19 − 2T must lie between −6 and 6, which forces T = 7 or T = 8.

8. T = 7 gives R − X = 5, satisfied only by R = 8, X = 3 among the remaining digits — but that would leave {2, 4, 6} for F, S and Y, and no two of those digits are consecutive, so F + 1 = S (from the ten-thousands column) could never hold. T = 7 is therefore impossible, forcing T = 8.

9. T = 8 gives R − X = 3. Two digit pairs satisfy this: (R, X) = (6, 3) or (7, 4). The first would leave {2, 4, 7} for F, S and Y — none consecutive, so F + 1 = S fails. The second leaves {2, 3, 6}, where 2 and 3 are consecutive, giving F = 2 and S = 3, with the one remaining digit, 6, going to Y. So R = 7, X = 4, F = 2, S = 3, Y = 6.

10. Collecting every assignment: F = 2, O = 9, R = 7, T = 8, Y = 6, E = 5, N = 0, S = 3, I = 1, X = 4 — all ten digits 0-9 used exactly once, each one forced by the column constraints rather than guessed.

Substituting back: FORTY = 29786, TEN = 850, and 850 + 850 + 29786 = 31486, which equals SIXTY = 31486 exactly — confirming the digit assignment is consistent.

So S = 3, I = 1 and X = 4, giving S × I × X = 3 × 1 × 4 = 12.