The secret code

From 2.) we have that $B = D + 1$ , and from 3.) we have that $E = 2D - 1$ .

So substituting these results, along with 1.), into 5.), we find that

$A + B + C + D + E = 30 \Longrightarrow 14 + (D + 1) + D + (2D - 1) = 30$

$\Longrightarrow 4D = 16 \Longrightarrow D = 4.$

So from 4.) we than have that $C = 10 - D = 6,$ and thus from 1.) we have $A = 14 - C = 8.$ From 2.) we have that $B = D + 1 = 5$ , and from 3.) we have $E = 2D - 1 = 7.$

Thus $ABCDE$ is the number $\boxed{85647}.$

Since each variable can be solved in terms of D except A, we need only find a substitution that allows us to solve for A in terms of D. We taken D + C = 10 and solve for C which yields C = 10 - D. We now substitute C in the first equation and get A - D = 4. Solving this last equation for A we get A = D + 4. We now have an equation for each variable in terms of D, and can now solve for D.

A = D + 4 B = D + 1 C = 10 - D D = D E = 2D - 1 A + B + C + D + E = 30

Now,

(A + 4) + (D + 1) + (2D - 1) + (10 - D) = 30 4D + 14 = 30 4D = 16 D = 4

We, now solve for the remaining variables:

A = D + 4, so A = 8 B = D + 1, so B = 5 C = 10 - D, so C = 6 D = 4 E = 2D - 1, so E = 7

Thus, ABCDE = 85647.

Step By Step For Beginners A + C = D + C + 4....... {From 1 &4}.............................................So A - D = 4 ................. A + B + C + D + E -D + D = 30 - D + D [Adding & Sub D] Rearranging A - D + B + C + D + E + D = 30 ........ [ On RHS (D-D) Can be neglected] 4 + B + C + D + E + D = 30 ............B + C + 2D + E = 26 .............2D = E +1...........[From 3]
B + C + E + 1 + E=26.......................B + C + 2E=25 ......D=10 - C [From 4] B - D=1............[From 2].............................................. B=11 - C [From above]................................................................. 11 - C + C + 2E=25..............2E = 14 .......E=7 ........D=4..........C=6...........A=8............B=5 So Finally ....A=8 B=5 C=6 D=4 E=7