
Vũ Thị Hương Giang 09/03/2017 at 22:30
We find that the sum of the digits of all the years before 1899 is less than 27, so the lucky year of those born before 1899 is before 1926. So the lucky year is after 1926. That person must be born from 1900 onwards.
Assume the year of birth is 19ab = 1900 + 10 a + b, the sum of the digits is 1 + 9 + a + b = a + b +10.
So the lucky year is: 19ab + a + b +10 = 1910 + 11a + 2b.
Similarly, if another person was born in 19cd, there was a lucky year of 1910 + 11c + 2d
For two years of equal luck, 1910 + 11a + 2b = 1910 + 11c + 2a or 11 (a  c) = 2 (d  b).
If
Db ≠ 0, so that the equality above satisfies (b  d) must be a multiple of 11, but this is absurd because b, d are digits. So d  b = 0, inference a  c = 0Then a = c; B = d. So they gave birth the same year.
Then year must find the lucky year of a person born in 19xy and a person born in 2abc.
Third Kamikare selected this answer. 
Trịnh Đức Phát 22/03/2017 at 12:51
We find that the sum of the digits of all the years before 1899 is less than 27, so the lucky year of those born before 1899 is before 1926. So the lucky year is after 1926. That person must be born from 1900 onwards.
Assume the year of birth is 19ab = 1900 + 10 a + b, the sum of the digits is 1 + 9 + a + b = a + b +10.
So the lucky year is: 19ab + a + b +10 = 1910 + 11a + 2b.
Similarly, if another person was born in 19cd, there was a lucky year of 1910 + 11c + 2d
For two years of equal luck, 1910 + 11a + 2b = 1910 + 11c + 2a or 11 (a  c) = 2 (d  b).
If
Db ≠ 0, so that the equality above satisfies (b  d) must be a multiple of 11, but this is absurd because b, d are digits. So d  b = 0, inference a  c = 0Then a = c; B = d. So they gave birth the same year.
Then year must find the lucky year of a person born in 19xy and a person born in 2abc.

FA FIFA Club World Cup 2018 16/01 at 22:04
We find that the sum of the digits of all the years before 1899 is less than 27, so the lucky year of those born before 1899 is before 1926. So the lucky year is after 1926. That person must be born from 1900 onwards.
Assume the year of birth is 19ab = 1900 + 10 a + b, the sum of the digits is 1 + 9 + a + b = a + b +10.
So the lucky year is: 19ab + a + b +10 = 1910 + 11a + 2b.
Similarly, if another person was born in 19cd, there was a lucky year of 1910 + 11c + 2d
For two years of equal luck, 1910 + 11a + 2b = 1910 + 11c + 2a or 11 (a  c) = 2 (d  b).
If
Db ≠ 0, so that the equality above satisfies (b  d) must be a multiple of 11, but this is absurd because b, d are digits. So d  b = 0, inference a  c = 0Then a = c; B = d. So they gave birth the same year.
Then year must find the lucky year of a person born in 19xy and a person born in 2abc.
We find that the sum of the digits of all the years before 1899 is less than 27, so the lucky year of those born before 1899 is before 1926. So the lucky year is after 1926. That person must be born from 1900 onwards.
Assume the year of birth is 19ab = 1900 + 10 a + b, the sum of the digits is 1 + 9 + a + b = a + b +10.
So the lucky year is: 19ab + a + b +10 = 1910 + 11a + 2b.
Similarly, if another person was born in 19cd, there was a lucky year of 1910 + 11c + 2d
For two years of equal luck, 1910 + 11a + 2b = 1910 + 11c + 2a or 11 (a  c) = 2 (d  b).
If
Db ≠ 0, so that the equality above satisfies (b  d) must be a multiple of 11, but this is absurd because b, d are digits. So d  b = 0, inference a  c = 0Then a = c; B = d. So they gave birth the same year.
Then year must find the lucky year of a person born in 19xy and a person born in 2abc.