THREE
 THREE
+ FIVE
------
ELEVEN

The leading E in the hundred-thousands column comes from a carry and can only be 1 or 2

From the ones column, E+E+E=N means N can only be 3 or 6, and there is no carry.

Now from the restriction we must have L+V+N-E-E-E = 0 mod 11.  From what we already know about N and E this congruence reduces to L+V = 0 mod 11.  Then from L and V being in 0 to 9 we can narrow this down to L+V = 11.

Next look at the tens column.  E+E+V=E then reduces to E+V=0, so then there is a carry of 1 and if E=1 or 2 then V=9 or 8

If E=2 and V=8 then L=3.  Then the ten-thousands column (carry)+T+T=23. T is at most 9 and the carry is at most 2 for a sum of 20, less than the 23 needed so this branch is a dead-end.

We must have E=1, V=9, L=2, and N=3.  Remaining digits are {0,4,5,6,7,8}. So far:

 THR11
 THR11
+ FI91
------
121913

A quick look at the thousands column has (carry)+H+H+F=1.  This must have a carry.

Looking at the ten-thousands column we have (carry)+T+T=12.  With two T's the carry can only be even, and we know there is a carry so it must be 2.  Then 2+T+T=12 means T=5.

Back to the thousands column. We know the carry into the ten-thousands column is 2 so (carry)+H+H+F=21.  0 and 4 are too small to be used for H or F, valid possible pairs for H and F are (carry,H,F)=(1,7,6), (2,6,7), or (1,6,8). Note 6 must be H or F.

The hundreds columns has (1)+R+R+I=9 or 19 or 29. But we know there is a carry to the thousands column and at the same time 29 is too large.  So 1+R+R+I=19. And R and I come from {0,4,7,8}.  The only way to get 19 is R=7 and I=4.

With R=7 then the thousands column can be revisited and only one of the three options remains, H=6 and F=8.  So final result is

 56711
 56711
+ 8491
------
121913