Letter Substitutions – 12
Rules:
- A letter represents a digit from 0 to 9, and has the same value throughout a single puzzle.
- No number can start with the digit 0.
- Within a puzzle, different letters must have different values.
THE CHALLENGE
Find the value of S, A, T, U, R, N, P, L, and E to make this puzzle work.
EXPLORATION
Make some letter substitution puzzles for others to solve.
Notes
THE CHALLENGE
Because carrying in adding problems with two numbers can be at most 1, we know that P must be 1. Also, N + S = S forces N = 0.
With those values, the puzzle becomes:
Looking at U + 0 + (possible carry) = E tells us that E is one more than U, there is a carry from the previous column, and there is no carry to the next column. That then means that T + A = 10, and there is a carry to the next column. A + R + (carry of 1) = 1A means that R is 9 and there is a carry to the next column.
We now know E = U + 1, T + A = 10, and the puzzle looks like this:
Looking at the tens column, 9 + U + (no carry) = 1T means that T is one less than U. Consequently, we have three digits in a row: T, U, and E.
Look through the possibilities using T + A = 10. The remaining question is how to make S + U + (carry) = 1L?
- T = 2, U = 3, E = 4, A = 8. Unused so far: 5, 6, 7. It’s not possible to solve S + 3 + 1 = 1L.
- T = 3, U = 4, E = 5, A = 7. Unused so far: 2, 6, 8. It’s not possible to solve S + 3 + 1 = 1L.
- T = 4, U = 5, E = 6, A = 6. Impossible with A = E.
- T = 5, U = 6, E = 7, A = 5. Impossible with A = T.
- T = 6, U = 7, E = 8, A = 4. Unused so far: 2, 3, 5. S = 5 and L = 3 works!
We end up with S = 5, A = 4, T = 6, U = 7, R = 9, N = 0, P = 1, L = 3, and E = 8. The solution looks like this:
