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.

  • $\begingroup$ Do you wish to solve this using the software Mathematica? $\endgroup$ – Manuel --Moe-- G Mar 18 '16 at 16:30
  • $\begingroup$ Using logic preferably. $\endgroup$ – Mathlete Mar 18 '16 at 16:30
  • $\begingroup$ This question will be moved, this here is for the software Mathematica. m0[n_] := If[n <= 365, Total[Table[i, {i, n, 365, n}]], 0] and Sort[{2016 - m0[#] === #, #, m0[#]} & /@ Select[Divisors[2016], # < 200 &]] solves the problem with brute force. Sorry I can't be more helpful. $\endgroup$ – Manuel --Moe-- G Mar 18 '16 at 16:38
  • $\begingroup$ No problem. It's just that in the maths competition we didn't even have a calculator. So there must be a logical way of doing it. $\endgroup$ – Mathlete Mar 18 '16 at 16:39

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 $4032$.

  • $\begingroup$ Clever solution. I just don't understand how you got the 4032 = x(n^2 - n + 2). Could you elaborate please. Thanks. $\endgroup$ – Tom Finet Mar 18 '16 at 21:13
  • $\begingroup$ $2016=x+x \sum_{i=1}^n i=x+x\frac{(n-1)n}{2}=x\frac{n^2-n+2}{2}$ $\endgroup$ – Nicolò Mar 18 '16 at 22:12

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