Is it possible to solve the following problem without make a "common sense" assumption?
In the year 1887, one person's age was exactly the sum of the digits of his birth's year. What was the person's age?
Just to be more clear: With common sense I mean, for example, not to assume that a man can't live more than 120 years, for example.
On
To avoid using too much "common sense" you could observe that the maximum digit sum of any year (CE) not later than 1887 is 27, so the subject was born after 1859.
Now the simplest way of proceeding is to tabulate the digit sums of years from 1860 to 1887 and the age of the subject in each of those years to deduce he was born in 1866 and is 21 years old
I'm not sure what you are referring to by "common sense" assumption. A common assumptions one would make with this problems is that the person age is within a certain range, say more than 0 and less than 100. Even without that assumption we can solve it. I think the only assumption we have to make is that his birth year was greater or equal to zero.
So let his birth year be $1000a+100b+100c+d$ with $0\leq a,b,c,d\leq9$ and $a,b,c,d\in\mathbb{Z}$. Hence:
$1000a+100b+10c+d+a+b+c+d=1887$
$1001a+101b+11c+2d=1887$
As $1001\times0+101\times9+11\times9+2\times9<1997$ then $a>0$.
As $1001\times2+101\times0+11\times0+2\times0>1997$ then $a<2$, hence $a=1$.
So $101b+11c+2d=886$
As $101\times7+11\times9+2\times<886$ then $b>7$.
As $101\times9+11\times0+2\times0>886$ then $b<9$, hence $b=8$.
So $11c+2d=78$
As $11\times5+2\times9<78$ then $c>5$.
As $11\times8+2\times0>78$ then $c<8$.
If $c=7$ then we require $11\times7+2\times d=78$ which gives $d=\frac{1}{2}$ which isn't allowed as $d$ is an integer.
So $c=6$ and hence $2d=12$ and hence $d=6$.
So the man was born in $1866$ and is $21$ years old.
Update: Even if we let him be born in BCE times (and represent this by a negative value for $a$) then above rational still works.