# A Problem from BdMO 2016 Regionals

Fifty natural numbers are written in such a way so that sum of any four consecutive numbers is $53$. The first number is $3$, the $19^\text{th}$ number is eight times the $13^\text{th}$ number, and the $28^\text{th}$ number is five times the $37^\text{th}$ number. Find the $44^\text{th}$ number.

I have nothing to say on this problem. How to solve this problem?

Hint: $3,a,b,(50-a-b),3,a,b,(50-a-b).....$ the pattern continues.
Hint: If the first five numbers are $v, w, x, y, z$, then you have the condition $v + (w + x + y) = 53$ and $(w + x + y) + z = 53$.
From $$a_1+a_2+a_3+a_4=a_2+a_3+a_4+a_5\Rightarrow a_1=a_5$$ we have similarly$$a_1=a_{1+4n}\\a_2=a_{2+4n}\\a_3=a_{3+4n}\\a_4=a_{4n+4}$$ Hence, because $13\equiv 37\equiv 1\pmod 4\Rightarrow a_{13}=a_{37}=a_1=3$ $$\begin{cases}a_{19}=8a_{13}\\ a_{28}=5a_{37}\end{cases}\Rightarrow\begin{cases}a_{19}=8\cdot3=24\\a_{28}=5\cdot3=15\end{cases}$$ It follows, because of $44\equiv28\pmod4$, $$\color{red}{a_{44}=a_{28}=15}$$