$2017-(a+b+c+d)=n$ where $n=abcd_{(10)}$ (Local Math Olympiad Qualification Exam Question)

Question

The problem text states: "'Person A' was born in 1962, in 1980 he turned 18 which is the sum of the numbers in his birth year (1+9+6+2). 'Person B' has a younger brother 'Person C', in 2017 the age of each one of them was equal to the sum of the digits in their birth year, when was 'Person B' born?". How can I approach this problem? I enumerated the cases using a script and found out that the possible numbers n that satisfy the requirements are 2012 and 1994.

Answer 1

The possible cases for $a$ are $a=1$ and $a=2$.

Case $a=2$.

We have $2017-2-b-c-d=2000+100b+10c+d$, that is, $$ 15=101b+11c+2d $$ Therefore $b=0$. Note that $c$ must be odd, so $c=1$ and $d=2$: the year is $2012$.

Case $a=1$

We have $2017-1-b-c-d=1000+100b+10c+d$, that is, $$ 1016=101b+11c+2d $$ Note that $11c+2d\le 99+18=117$, so we need $101b\ge899$, that is, $b\ge9$. Hence $b=9$ and $11c+2d=107$. Since $2d\le18$, we have $11c\ge89$, so $c\ge9$, hence $c=9$ and $d=4$. The year is 1994.

Comment

The younger brother is needed, or person B could have been born either in 1994 or 2012.

Answer 2

If you want do to this without a script.

Born in 2010 - 2017.

$7-d = 3+d\\ d = 5$

Born in 2000 - 2009.

$17 - d = 2+d\\ 2d = 15\\ d = 7\frac 12$

If you are looking for integer solutions, none exist. But, as you don't turn your age until your birthday comes, born in $2007$ can work.

Born in 1990 - 1999.

$27-d = 19+d\\ d = 4$

Born in 1980 - 1989.

$37-d = 18+d\\ d = 9\frac 12$

$1989$ is works depending on your rules.

Your age must be less than $1+9+9+9 = 28$

We could perhaps generalize it a little more and say

$17 - 10c - d = 2 + c + d\\ 9c + 2d = 15$

or

$117 - 10c - d = 10 + c + d\\ 9c + 2d = 107$

If we want $c, d$ to be in $\{0,1,2,3,4,5,6,7,8,9\}, c $ must be odd and sufficiently small in one equation and large in the other.