Means A+A=E, N+V=K, A+A=N, N+U=A, A+K=R, M+_=S

So just start from last

A+A=E

"A" can have d values ranging {0,1,2,3,4,5,6,7,8,9}

First let’s take A=0... A+A= 0+0=0... Again d result is 0. Means E=0... So not possible for 2 alphabets to have same number... 

Let’s take A=1, A+A= 1+1=2... A=1, E=2 ... v have to find d value for "N". 

Look A+A=N (3rd term), A+A is already "E" (ie E=2A). So, definitely N should not be 2A... Therefore N=2A+1, there should be remainder from the 2nd term.

So N=2A+1. N=2(1) +1= 3. Assign a value for to get a remainder. Take V=7

3+7=0 carry 1.

1+A+A=3... 

Next N+U=A, result should be 1 

=>3+8=1 carry 1... 

Next A+K=R... 1+1+7=9, there is no carry which means B and S will have d same value... so not possible... So try for different values...

Note here A+K=R should produce a carry... then only v can yield M + carry=S

By trial and error method, again start from last. Try A=2... It’s also not possible... Again A=3. Not possible.

Now A=4, E=2A=8... N=2A+1=9. 

For getting a carry If we take V=1, A+K will not produce carry. So, take V=2, N+V=K

9+2=1 carry 1, 1+A+A=9. N+U should produce 4. 9+5=4 so U=5.

Next A+K = 4+1=5 not producing carry... Here 4+ (>5) only yields a carry.

So v should get K value >5.

So again trail and error method, N+V=K. Let’s take V=7, N+V=K, 9+7=6 carry 1.

1+A+A=N, N+U=A, 9+5=4 carry 1.

1+A+K=R... 1+4+6=1 carry 1... 1+B+_=S... 

Values remaining is {0, 2, 3}

1+0+_=1 not possible

1+2+_=3

So, 

M=2, A=4, N=9, A=4, N=9, A=4

K=6, U=5, A=4, V=7, A=4

S=3, R=1, A=4, N=9, K=6, E=8

S + R + A + N + K + E 

3 + 1 + 4 + 9 + 6 + 8 = 31