magic cryptarithm

  HG GH UF LI  U
  IF  I EU GG UH
  NU UG IH FF EI
  FH HF NI RU IG
  RI LU FG HH GF

 This is a magic square with magic constant RHH.
 The 25 numbers form an arithmetic progression.
 If you order the letters according there values you get the hidden word.




Solution:
The two digit numbers have nine different first digits.
Therefore the stepsize must be 4. What is the starting value
of the arithmetic progression? 

  1,  5,  9, 13, ...,  97  first three numbers have one digit.
  2,  6, 10, 14, ...,  98  the digit 0 would need a tenth letter.
  3,  7, 11, 15, ...,  99  only case not excluded.
  4,  8, 12, 16, ..., 100  last term too big.

Now calculate the magic constant. The sum of all numbers is

  s =  3 +  7 + 11 + ... + 95 + 99
  s = 99 + 95 + 91 + ... +  7 +  3
 ---------------------------------
 2s =102 +102 +102 + ... +102 +102  =  25*102

Therefore s = 25*102/2 = 1275
The magic constant is m = s/5 = 255.
So we know R=2 and H=5.
Further we have the one digit numbers {I,U} = {3,7}.
The other odd digits must be {F,G} = {1,9}.
So we have for the even digits {E,L,N,R} = {2,4,6,8}.
Now separate the two digits in two squares

  HG GH UF LI  U          H G U L 0     G H F I U
  IF  I EU GG UH          I 0 E G U     F I U G H
  NU UG IH FF EI  =  10 * N U I F E  +  U G H F I
  FH HF NI RU IG          F H N R I     H F I U G
  RI LU FG HH GF          R L F H G     I U G H F

The last digit square is an Euler Square. Each row, column,
and diagonal contains the letters F,G,I,H,U once.
So it has the magic constant m1 = 1+3+5+7+9 = 25.
As the linaer combination of magic squares is a magic too
we get that the first digit square is magic with constant m2 = 23.
This is an implication of m = 10*m2 + m1.
Now solve the linear equations for the first digit square.
(C4)  1+5+9+L+R = m2  =>  L + R = 8 
Therefore {L,R} = {2,6} and the complement {E,N} = {4,8}.
(R2)  3+7+E+G = m2  =>  E + G = 13
Thus E=4 and G=9. Further N=8 and F=1.
All other rows, colums, and diagonals only show that I+R = U+L = 9.
Therefore we have two possible solutions
 I=5+t, U=5-t, L=4+t, R=4-t  with t in {-2, 2}.
The linear system was underdetermined. Both solutions are give an
arithmetic progression. (Modulo 4 the variables I,U,L,R are the same
in both solutions.)

  t = -2              t = 2

 59 95 71 23  7      59 95 31 67  3
 31  3 47 99 75      71  7 43 99 35
 87 79 35 11 43      83 39 75 11 47
 15 51 83 67 39      15 51 87 23 79
 63 27 19 55 91      27 63 19 55 91

 123456789           123456789
 FLIEHRUNG           FRUEHLING

Only the second solution has a valid german word. It means spring.

References:

Zweisteins Zahlenlogeleien,
  Insel Tb. 1510, 1993, ISBN 3-458-33210-3
  Problem 22