25) F

Let the first digit be F
Let the third digit be T
Let the fourth digit be M
Let the fifth digit be N
Let the sixth digit be S
We know the second digit is 8.

               F8
          +  TM
               NS
                80

              F8T
          + MNS
               800

REMEMBER: All digits are less than 10, meaning that the sum of two digits cannot be higher than 17. (19 = 9 + 10), (18 = 9 + 9 but no two digits can be same)

By looking at the second addition, we can see that T + S = 10
Ways to make 10 = (7+3), (6+4). It cannot be 8+2 as 8 is already a digit and it cannot be 5+5 as no two digits can be the same. The 1 from the 10 will be carried over in the sum 8 + N.
Since 8 + 1 + N = 10, N = 1

The 1 will be carried over to F + M.
F + M + 1 = 8
F + M = 7
This can only be either 4 + 3, 5 + 2. It cannot be 6 + 1 as N = 1.

Going to the first sum, 8 + M + S = 20 (it cannot be 10 or 30 or more).
M + S = 12
This can only be 7 + 5

The 2 from 20 will be carried over
F + T + N + 2 = 8 (N=1)
F + T = 5

Possibilities of T = 7, 3, 6, 5.
Since F + T = 5, then  T < 5
So T is either 4 or 3.
If T = 4, then F = 1 (which cannot be true as N = 1)
So T = 3.
F = 5 – 3 = 2

F + M = 7
M = 5

M + S = 12
S = 7

The passcode is 2 – 8 – 3 – 5 – 1 – 7

The last digit of passcode = S = 7

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