$11c+2d=86$ has no solution.

Question

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...

Answer 1

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.