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