Problem: Alan Wayne

  E I N  +  E I N  +  E I N  +  E I N  =  V I E R





Solution:

Select the two digits E and I.

  4 * (10E + I) + c = 10I + E  (mod 100)

We are looking for solutions with a carry 0<=c<4.
Simplification gives

  39*E + c = 6*I  (mod 100)

Create a table trying all different values of E.
If E is fixed the values of I and c can be determined.

 E    39*E    39*E (mod 100)   I    6*I     c     good of bad?
 0       0       0             0      0     0      E = I
 1      39      39             7     42     3      good
 2      78      78            13     78     0      I > 9
 3     117      17             3     18     1      E = I
 4     156      56            10     60     4      I > 9, c > 3
 5     195      95            16     96     1      I > 9
 6     234      34             6     36     2      E = I
 7     273      73            13     78     5      I > 9, c > 3
 8     312      12             2     12     0      good
 9     351      51             9     54     3      E = I

Now check the two good candidates.
case E=1, I=7, c=3:
  We get V=0 so we have a leading zero.
  To get c=3 we must have 4*N >= 30.
  Therefore we can have N=8 or N=9.
  This gives the solutions
  4 * 178 = 0712 and 4 * 179 = 0716.
case E=8, I=2, c=0:
  We get V=3.
  To get c=0 we must have 4*N < 10.
  Therefore we have N = 1, as 2 is used already and 0 would give R=0 too.
  This gives the solution
  4 * 821 = 3284.

Reference:
   Martin Gardner, Scientific American (Dec. 1975), p116