Kim was born on first January many years ago and in 2002 her age is the sum of all the four digits of the year she was born in how old was she in 2002?

Answer 1

Let's say she was born in abcd

The statement says

2002-(1000a+100b+10c+d)=a+b+c+d

a+b+c+d+(1000a+100b+10c+d)=2002

1001a+101b+11c+2d=2002

If a=2 there's one solution: 2000

Then if a=1

As for b, if b=8 then Kim is born between 1800 and 1899 and the sum a+b+c+d should be more than 103 which is impossible with 4 digits.

So when can take b=9 and suppose that Kim is born between 1900 and 1999.

1001a+101b+11c+2d=2002 becomes 1*1001+9x101+11c+2d=2002

then 11c+2d=92

As 11 and 2 are primes, they are coprime

and one finds easily that 11x1-2x5=1 so 11x(1-2k)-2(5-11k)=1 for k in N

then 92 [ 11x(1-2k) - 2x(5-11k) ] = 92

or 11x(92-184k) - 2x(460-1012k) = 92

c=91-184k and d=1012k-460

and c and d are between 0 and 9 (as digits)

there's no such k

So the only solution is 2000