application and triangulation of filters, which were more faithful to each other were the diagonal I call 1-48, 6-49 and diagonal endings that then I will explain who is who. However it is important to say that the filter of the terminations, works as a filter applied within the diagonal 1-48 since each diagonal of 1-48 has only two endings.
Now let's see: in this image we have 13 diagonals that compose the diagonals of 1-48, I baptized it from 1-48 because starts at 1 and ends at 48.
If noticed, each diagonal of these contains only 2 endings.
Diag. 1-01
Diag. 2-02.07
Diag. 3-03.08.13
Diag. 4-04.09.14.19
Diag. 5-05.10.15.20.25
Diag. 6-06.11.16.21.26.31
Diag. 7-12.17.22.27.32.37
Diag. 8-18.23.28.33.38.43
Diag. 9-24.29.34.39.44
Diag. 10-30.35.40.45
Diag. 11-36.41.46
Diag. 12-42.47
Diag. 13-48
As for the Diagonals of 6-49 have 14 diagonals that compose it, also I baptized her 6-49 because starts at 6 and ends at 49. If we analyze these diagonal endings, also can fix that in each there is a single ending that repeats itself.
And each diagonal of 6-49 is composed of:
Diag. 1-06
Diag. 2-05.12
Diag. 3-04.11.18
Diag. 4-03.10.17.24
Diag. 5-02.09.16.29.30
Diag. 6-01.08.15.22.29.36
Diag. 7-07.14.21.28.35.42
Diag. 8-13.20.27.34.41.48
Diag. 9-19.26.33.40.47
Diag. 10-25.32.39.46
Diag. 11-31.38.45
Diag. 12-37.44
Diag. 13-43.50
Diag. 14-49
Now, bearing in mind that in the Euromillions only leave 5 balls, on top, just that choice also only 5 diagonals to play, both in 1-48 as in 6-49, and 5 endings, not obvious? When doing this application will almost always reduced numbers of a quantity of 5 to 11 final numbers. But also, if the choices are "poorly made" we can having a reduction of less than 5 figures, which will make it possible to add some more choices.
However, the problem lies in that the diagonals and terminations to choose! As for this problem, by recall and a bit of luck (just a little!)...
For example ... and only as an example!
If we choose in diag. 1-48 the following 5 diagonals:
Diag. 3-03.08.13
Diag. 5-05.10.15.20.25
Diag. 6-06.11.16.21.26.31
Diag. 8-18.23.28.33.38.43
Diag. 11-36.41.46
And diag. 6-49 as follows:
Diag. 4-03.10.17.24
Diag. 6-01.08.15.22.29.36
Diag. 8-13.20.27.34.41.48
Diag. 9-19.26.33.40.47
Diag. 11-31.38.45
As you can see, combining filters, we were only with a total of 12 numbers, which are put in bold (03.08.10.13.15.20.26.31.33.36.38 and 41)
Now, when we apply the filter of terminations if we choose 0 endings, 1, 3, 6 and 8 we get a final result of 11 numbers:
After we get to the final result, we must first of all, look at the