Letter Substitutions – 13
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 P, O, T, A, M, L, E, U, and C 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 L must be 1.
Note that in the ones column O + O = E and that in the ten thousands column O + O + (possible carry) = T. So the “possible carry” is a definite carry. Next notice one column to the right that we have T + M + (possible carry) = 1T. If M is 0, the sum will be less than 10. Therefore, M = 9 and the possible carry is a definite carry. Note also that O + O = E and O + O + 1 = T forces T = E + 1.
With those values, the puzzle becomes
P + T + (possible carry) = E and T = E + 1 forces P to be 8 or 9. However, 9 is taken. So P = 8 and, once again, the possible carry is a definite carry. Given the carry information we have established, we also know that O + O is at least 10 and A + A is at least 10. Put another way, O and A are at least 5.
To summarize, here is our updated puzzle, and we also know T = E + 1, C = T + T + 1, and O and A are at least 5
At this point, we have 0, 2, 3, 4, 5, 6, and 7 to work with, and we need to find the value of E, T, C, O, A and U. Note that O + O = E forces E to be even.
- E = 2, T = 3, C = 7, O = 6.
- E = 4, T = 5, C = 1, O = 7. This is impossible as it forces C = L = 1.
- E = 6, T = 7, C = 5, O = 8. This is impossible as it forces O = P = 8.
Therefore E = 2, T = 3, C = 7, O = 6, P = 8, M = 9, and L = 1. That only leaves 0, 4, and 5 for A and U. Fortunately, A = 5 and U = 0 works, and we are done! Here is the finished puzzle
