We are in the year $2016$, and Cedric's age is a factor of $2016$. If Cedric adds up all the multiples of his age that are less than $365$, he arrives at the year he was born.
In which year was he born?
I have tried this question numerous times and have not been successful.
Let $x$ be the age and $n$ the greatest natural such that $xn \le 365$. Then the equation is $$ 2016-x=\sum _{i=1}^n xi $$ $$2016=x+ x\sum _{i=1}^n i $$ $$ 2^6 \cdot 3^3 \cdot 7=2\cdot 2016=4032=x(n^2-n+2) $$ Now, $xn^2-xn+2x\le 365n-365+2x \le 365(n-1) $ and since $365 \cdot7 >4032$ it is necessary that $n-1 \le 7$. The only possible values for $n$ are up to $8$.
Keeping in mind that $x \le 365$ using $4032=x(n^2-n+2)$ you exclude $n=1,2$ and knowing that $n^2-n+2$ divides $4032$ you only get the possible results of $n=3,4,8$. Just check the corresponding $x=\frac{4032}{n^2-n+2}$ and see for which one $xn\le 365$ holds. They are all easy computations if you use the factorization of $4032$.