$11c+2d=86$ has no solution. 
Show that, in the year $1996$, no one could claim on his birthday, that his age was 
  the sum of the digits of the year, in which, he was born.

My attempt:-
 Suppose a person born in the year $19\bar{cd}$ claims that his age ,in $1996$ , is equal to the sum of the digits of $19\bar{cd}$. This implies that $$1996-19\bar{cd}=1+9+c+d$$
Therefore, We can write $$96-\bar{cd}=10+c+d$$ 
This implies that $$96-10c-d=10+c+d$$
Which reduces to  $$11c+2d=86$$ It is easily checked that no $c,d$ satisfying $0 \le c,d\le9$ satisfy the above linear diophantine equation. Hence the result. Does this look correct?(I think ,not) .
P.S. We do not consider the years $18..$ ,$17..$ etc. As the largest sum(of digits) possible for any four digit number is 36...

Using trial and error , one can find that (0,43) are the solutions of the given diophantine equation.
Hence the general values of  x and y are:
$$C = 2t $$
$$D = 43-11t$$
for positive values of x and y :
$$t< 4$$
and the possible pairs of positive c and d are $(2,32),(4,21)$ and $(6,10).$
But Since c,d < 10 (They Are Digits), There Is no possible solution.
