Daily Math Puzzle: 2026-07-22

19 — Let's break down the cryptarithm `TECH * 2 = POWER` step-by-step, where each letter is a unique digit (0-9), and T, P cannot be 0. 1. **Analyze the thousands column (and beyond):** Since `TECH` is a 4-digit number and `POWER` is a 5-digit number, multiplying a 4-digit number by 2 can at most result in `9999 * 2 = 19998`. This means the leading digit 'P' in `POWER` must be 1. So, **P = 1**. 2. **Analyze the thousands column (continued):** Let `c1, c2, c3, c4` be the carries from right to left (units to tens, tens to hundreds, hundreds to thousands, thousands to ten thousands). The multiplication equation can be written as: - Units: `2 * H = R + 10 * c1` - Tens: `2 * C + c1 = E + 10 * c2` - Hundreds: `2 * E + c2 = W + 10 * c3` - Thousands: `2 * T + c3 = O + 10 * c4` - Ten thousands: `c4 = P` From `c4 = P` and `P = 1`, we get **c4 = 1**. 3. **Deduce 'T' and 'O':** From `2 * T + c3 = O + 10 * c4`, substitute `c4 = 1`: `2 * T + c3 = O + 10` Since `T` cannot be 0 and `T` cannot be 1 (as `P=1` and all letters are unique), `T` must be at least 2. `c3` can be 0 or 1. `2 * T + c3` must be at least 10 for `c4` to be 1. - If `T = 2`, `2*2 + c3 = O + 10` -> `4 + c3 = O + 10`. This implies `O` would be negative, which is impossible. - If `T = 3`, `2*3 + c3 = O + 10` -> `6 + c3 = O + 10`. Impossible. - If `T = 4`, `2*4 + c3 = O + 10` -> `8 + c3 = O + 10`. Impossible. So, `T` must be at least 5. Let's test `T >= 5`: - If `c3 = 0`: `2 * T = O + 10`. Possible values for (T, O): - (5, 0) `(2*5 = 10 => O=0)` - (6, 2) `(2*6 = 12 => O=2)` - (7, 4) - (8, 6) - (9, 8) - If `c3 = 1`: `2 * T + 1 = O + 10`. Possible values for (T, O): - (5, 1) `(2*5+1=11 => O=1)`. But `O` cannot be `P` (which is 1). So (5,1) is not possible. - (6, 3) - (7, 5) - (8, 7) - (9, 9) `(2*9+1=19 => O=9)`. But `O` cannot be `T` (which is 9). So (9,9) is not possible. 4. **Analyze `2 * E + c2 = W + 10 * c3`:** We need to determine `c3` (0 or 1). Also, `E` cannot be `P=1`. `W` cannot be `P=1`. 5. **Let's assume `c3 = 0` (from `(T, O) = (5, 0)`):** - From `2 * T + c3 = O + 10`, if `c3 = 0`, then `2 * T = O + 10`. The only allowed unique pair not using `P=1` is `T=5, O=0`. So, **T = 5, O = 0**. - Now, `2 * E + c2 = W + 10 * c3` becomes `2 * E + c2 = W + 10 * 0` => `2 * E + c2 = W`. (This implies `c2` must be 0 for `W` to be a single digit, and for `2E+c2` to not be >9. Let's check). - If `c2 = 0`: `2 * E = W`. `E` cannot be 0, 1, 5. Possible `E, W` pairs: (2, 4), (3, 6), (4, 8). - If `c2 = 1`: `2 * E + 1 = W`. `E` cannot be 0, 1, 5. Possible `E, W` pairs: (2, 5) (but `W=5=T`, so no), (3, 7), (4, 9). 6. **Analyze `2 * C + c1 = E + 10 * c2`:** - Let's try `c2 = 1` from the `2 * E + c2 = W` step. So `2 * E + 1 = W`. Let's pick `E=3, W=7`. (Used: P=1, T=5, O=0, E=3, W=7). - Now: `2 * C + c1 = E + 10 * c2` becomes `2 * C + c1 = 3 + 10 * 1` => `2 * C + c1 = 13`. - `C` cannot be 0, 1, 3, 5, 7. So `C` can be `2, 4, 6, 8, 9`. - If `c1 = 0`: `2 * C = 13` (not possible, `C` is integer). - If `c1 = 1`: `2 * C + 1 = 13` => `2 * C = 12` => **C = 6**. (Used: P=1, T=5, O=0, E=3, W=7, C=6). 7. **Analyze `2 * H = R + 10 * c1`:** - We found `c1 = 1`. So, `2 * H = R + 10`. This means `H` must be >= 5. - `H` cannot be 0, 1, 3, 5, 6, 7. So `H` can be `2, 4, 8, 9`. - Since `H >= 5`, `H` must be `8` or `9`. - If `H = 8`: `2 * 8 = R + 10` => `16 = R + 10` => `R = 6`. But `R` cannot be `C` (which is 6). So `H` is not 8. - If `H = 9`: `2 * 9 = R + 10` => `18 = R + 10` => **R = 8**. (Used: P=1, T=5, O=0, E=3, W=7, C=6, H=9, R=8). 8. **Verify the solution:** - P=1, O=0, W=7, E=3, R=8 - T=5, C=6, H=9 All digits {0, 1, 3, 5, 6, 7, 8, 9} are unique and `T, P` are not zero. This is a valid solution. Let's check the multiplication: `TECH = 5369` `POWER = 10738` `5369 * 2 = 10738`. This is correct. 9. **Answer the question:** The value of 'POWER' is 10738. The sum of the digits that spell out 'POWER' is `P + O + W + E + R = 1 + 0 + 7 + 3 + 8 = 19`. The final answer is **19**.