Problem source

A secret message consists of the numbers $1,3,7,23,24,37,39,43,43,43,45,47$ arranged in some order as $a_{1},a_{2},\ldots,a_{12}.$ The message is encoded as $b_{1},b_{2},\ldots,b_{12}$ with $0\leqslant b_{j}\leqslant49$ and 
\begin{alignat*}{1}
b_{2j} & \equiv a_{2j}+n_{0}+j\pmod{50},\\
b_{2j+1} & \equiv a_{2j+1}+n_{1}+j\pmod{50},
\end{alignat*}
for some integers $n_{0}$ and $n_{1}.$ If the coded message is $35,27,2,36,15,35,8,40,40,37,24,48,$
find the original message, explaining your method carefully. 

Solution source

Considering the odd numbers, we have

\begin{array}{l|rrrrrr}
b_{2j+1} & 35 & 2 & 15 & 8 & 40 & 24 \\
a_{2j+1}+n_1 & 35 & 1 & 13 & 5 & 36 & 19 
\end{array}

Considering the even numbers, we have

\begin{array}{l|rrrrrr}
b_{2j} & 27& 36 & 35 & 40 & 37 & 48 \\
a_{2j}+n_0 & 27 & 35 & 33 & 37 & 33 & 43 
\end{array}

There are three numbers in the original sequence which are repeated ($43$). By the pigeonhole principle, one of the odds or evens must have at least two of them. We can see that the even numbers have some number repeated twice ($33$). Therefore these must be the $43$s. Therefore $n_0 = -10$

\begin{array}{l|rrrrrr}
b_{2j} & 27& 36 & 35 & 40 & 37 & 48 \\
a_{2j}+n_0 & 27 & 35 & 33 & 37 & 33 & 43 \\
a_{2j} & 37 & 45 & 43 & 47 & 43 & 3
\end{array}

This leaves the remaining numbers to be decoded from the original sequence as $1,7,23,24,39,43$. Two of these numbers are consecutive ($23$ and $24$), and two numbers in our sequence are $35$ and $36$. Therefore $n_1$ must be $12$.

\begin{array}{l|rrrrrr}
b_{2j+1} & 35 & 2 & 15 & 8 & 40 & 24 \\
a_{2j+1}+n_1 & 35 & 1 & 13 & 5 & 36 & 19 \\
a_{2j+1} & 23 & 39 & 1 &43 & 24& 7
\end{array}


Therefore the original sequence was: $23, 37, 39, 45, 1, 43, 43, 47, 24, 43, 7, 3$