Her age on the 22 April, 2003 is 25 as she was born in 1978. If the year that she was born is abcd then we have the following age = 2003 - (a+b+c+d) In this case we will infer that she was born in the 20th century, therefore a = 1 and b = 9. age = 1993 - (c+d) from this stage i don't know where to go.

I don't think this is possible becasue you did not give us teh year she was born in or how old she is. Masoud Zargar

Quote: from this stage i don't know where to go.

Could you please tell me what "flanders" means? I guess it is a math contest..

LOL \[\left\{\textrm{Belgium}\right\}= \left\{\textrm{Flanders}\right\}\cap\left\{\textrm{Wallonia}\right\}\] (Belgium is a country, by the way.)

Now that I look at my post from over a year ago I feel ashamed.

Hmmm... So actually, Kurt Gödel, you mean: \[\left\{\textrm{Belgium}\right\}\ =\ \emptyset\]

Sorry `bout that..but usually when I see a thing in the form "..blabla.. `03" I think of a contest[especially because this is a math forum) And Kurt Godel, I guess you mean reunion, not intersection And oh..I knew it was a country,lol

Flanders is the region organising the contest. Just like some people say China '03 instead of CMO '03.

ya i did it similar to GYAN..except the $\mod$ part i just did some educated guessing and checking..looking at GYAN's solution i think its $2d\equiv{5}\mod{11}\Rightarrow{d}\equiv{8\mod11\therefore{d}=8}$ and the rest follows...

It is obvious that she is born in the year 19ab. Thus,her age is 10+a+b (1) Also,her age is also 2003-(1900+10a+b)=103-10a-b (2) From (1) and (2),93=11a+2b. Easy trial and error shows that a=7 and b=8. Hence she was born in 1978 and she is 25 years old.

Let the year she was born in be $\overline{abcd}$. Then we have $a, b, c, d \in \mathbb{N}$ such that \begin{align*} 2003 - (1000a + 100b + 10c + d) &= a + b + c + d \\ \implies 1001a + 101b + 11c + 2d &= 2003 \end{align*}Clearly, $a=1$ or $2$. If $a=2$, we have $101b+11c+2d=1$. If $b,c,d$ are all $0$, then $101b+11c+2d=0 \neq 1$. Hence at least one of $b,c,d$ is greater than $0$. We have $101b+11c+2d \geq 101(0)+11(0)+2(1) = 2 > 1$, a contradiction. Hence $a=1$ which gives \[101b+11c+2d = 2003-1001=1002\]Now note that Gittes' age $\leq 9+9+9+9=36 \implies \overline{abcd} \geq1967$. Since $a=1$, $b=9$, we have \[11c+2d=1002-101(9)=93\]Since $d\geq 9$, $93=11c+2d\geq 11c+2(9)=11c+18 \implies 11c \geq 75 \implies c \geq 7$. Additionally, $11c+2d\leq 93 \implies 11c \leq 93 \implies c \leq 8$. Hence $c=7$ or $8$. If $c=8$, $2d=93-11(8)=5 \implies d=\frac{5}{2}$, contradicting the fact that $d$ is an integer. Hence $c=7 \implies d=8$, so Gittes is born in $1978$.