Alright, here goes. 10 letters. 10 numbers.

First thing I notice. 6 Ts in first column. T must be 1, or else N > 10. N, therefore, must be 6 + the carry over from the last column. To be determined. 5 Vs Plus T is T, so the carry over from the previous column must either be 5 or 0. 5 Es plus Y is Y, so E cannot be odd. If E is not odd, then 5 Es plus Y cannot total 50 or more. Therefore, E must be 0 and V must be an even number. 5 0s and Y obviously will equal Y, so Y will be the last number to find. That said, I glanced at the E - I column. 0 plus I plus the carry over from the last line equals N. Since N > 6, there is no way for I to be a high number, so the carry over from that line must be 0. Given this, W must be odd, else 5 of them would make a 0, and 0 plus H cannot equal I. Since I must be a low number, as mentioned, H, therefore, must now be > 6. Since H > 6, W can be no greater than 5, as that would push the final column above 10. W must be 3 or 5. N must, therefore, be 8 or 9. If N is at least 8, then I must be at least 3, as the greatest carryover possible in a column is 5. Since I is less than 5 but greater than 2, that means it can only be 3 or 4. As a result, H can only be 8 or 9. That brings me to Ls and Rs. Since I is only 3 or 4, and N is 8 or 9, 5 Ls and an R must total 40 (as 50 or 60 is not possible due to taken slots). This makes L have to be 6 or 7. If L is 6, that gives 30 of 40. To achieve 10 more, R must be as high as possible, 7, but even then, V would be 4, and the total would be too low. Therefore, L has to be 7. Since we now know that this column totals 40, that means I + 4 = N. The only way to satisfy this is 4 and 8. So we get that I is 4, N is 8, and H is 9. Because of this, 5Ws plus H cannot be more than 20, so W must be 3. R can no longer be 6, so it must be 2, making V 6. Y, therefore, is 5.

T - 1

W - 3

E - 0

L - 7

V - 6

H - 9

I - 4

R - 2

Y - 5

N - 8

```
130760
130760
130760
130760
130760
194215
______
848015
```

tl;dr - I win