The partial order method [H. MüllerMerbach]
A very efficient method to solve number puzzles is to derive
as many "less"conditions as possible and connect them in the
form of a graph. Through systematically doing so, the values
of the variables can be found almost automatically. One may
switch to linear discursion after the first values have been
detected.
This technique will be demonstrated by the following example
(taken from Hör Zu 17/1972):
(I) ABCD : EFG = HC
 * +
(II) FHA  IG = FJD

(III) AHBE  ACFE = FAJ
(IV) (V) (VI)
"Less"conditins can be easily derived from the leading
letters of the single equations as follows:
From eq. (III) follows: J = 0
From eq. (VI) and J=0 follows: A = H+1
From eq. (II) and J=0 follows: I = H1
From eq. (III) follows: C < H and F < H
From eq. (IV) follows: F < B and H < B
From eq. (I) follows: E < A and H < A
From eq. (II) and J=0 follows: A < D and A < G
These relations gives the partial order
10
/  \
B D G
\  /
A

H

I
/  \
C E F
\  /
J = 0
From this partial order (graph) it follows immdiately:
I = 4, H = 5, and A = 6.
These values can now be used to find the values of all the
other letters by linear discursion: E = 1 (I); D = 7 (IV);
C = 3 (VI); G = 9 (II); F = 2 (III); B = 8 (III).
6837 : 129 = 53
 * +
256  49 = 207

6581  6321 = 260
This method of structuring a problem by means of a graph has
the advantage over linear and branched discursion that the
relations between the letters become very clear. It can very
quickly be carried out since the 6 equations can be examined
one after another. The knowledge of some basic algebraic rules
is sufficient.
References:
H. MüllerMerbach;
The Role of Puzzles in Teaching Combinatorial Programming,
In: B. Roy (ed)., Combinatorial Programming: Methods and
Applications, 379386, D. Reidel Publ. 1975