Suppose we call the digits ABCD. We have

D = 2*C

B = D - 5

A = C + B

Expressing these in terms of C, we have

B = 2*C - 5

A = C + 2*C - 5 = 3*C - 5

We know every digit must be in the range 0 to 9, so we have

9 ≥ 2C - 5 ≥ 0

14 ≥ 2C ≥ 5 (add 5)

7 ≥ C ≥ 2.5 (divide by 2)

9 ≥ 3C - 5 ≥ 0

14 ≥ 3C ≥ 5 (add 5)

4 2/3 ≥ C ≥ 1 2/3 (divide by 3)

In order to satisfy both these conditions, we must have C = {3, 4}. This gives rise to two possible solutions. For C=3, we have D=6, B=1, A=4. For C=4, we have D=8, B=3, A=7. The possible solutions are

D = 2*C

B = D - 5

A = C + B

Expressing these in terms of C, we have

B = 2*C - 5

A = C + 2*C - 5 = 3*C - 5

We know every digit must be in the range 0 to 9, so we have

9 ≥ 2C - 5 ≥ 0

14 ≥ 2C ≥ 5 (add 5)

7 ≥ C ≥ 2.5 (divide by 2)

9 ≥ 3C - 5 ≥ 0

14 ≥ 3C ≥ 5 (add 5)

4 2/3 ≥ C ≥ 1 2/3 (divide by 3)

In order to satisfy both these conditions, we must have C = {3, 4}. This gives rise to two possible solutions. For C=3, we have D=6, B=1, A=4. For C=4, we have D=8, B=3, A=7. The possible solutions are

**4136****7348**